Thought experiment: I acquired two boxes of the same dimensions and same weight. One box contains $1\ \mathrm{kg}$ of water at room temperature while the other box has $1\ \mathrm{kg}$ of water, but in steam form, because the temperature of the box is above $100^\circ\mathrm{C}$. The volume of the boxes is large relative to the amount of space the $1\ \mathrm{kg}$ of water would take (let's arbitrarily say $10\ \mathrm{L}$). Both boxes contain the same amount of air (at $1\ \mathrm{atm}$) which is why the second box has water in steam form at $100^\circ\mathrm{C}$.

I put each box on a simple electronic scale to measure their respective weights. Unsurprisingly, the box containing water comes out to be $1\ \mathrm{kg}$. But what about the box containing steam?

My guess: Electronic scales measure the amount of force being exerted on it, then divide that force by $g$, to get the mass of the object. I think the box with steam in it will be exerting less force on to the scale and therefore the scale will think its mass is less than $1\ \mathrm{kg}$.

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    $\begingroup$ You should clarify: are you asking about the same volume of the same mass of liquid and gas? $\endgroup$ – coconut Feb 28 '17 at 9:18
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    $\begingroup$ You may like to take a look at this question which explains why gases have weight in the first place. $\endgroup$ – BLAZE Feb 28 '17 at 9:19
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    $\begingroup$ "1L" does not specify an amount of matter: It specifies a volume. When you say, "1L of water, but in steam form," that sounds as if you are talking about a certain mass of water, but if that's what you mean, then you should describe it with a mass unit, not a volume unit. $\endgroup$ – Solomon Slow Feb 28 '17 at 18:03
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    $\begingroup$ The question from the OP needs to be reformulated. It's physically impossible to put 1 kg of steam into a l liter box, because the density of steam is below 1000 kg/m^3, even at the critical temperature and critical pressure. $\endgroup$ – David White Feb 28 '17 at 22:47
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    $\begingroup$ With edits the question is still flawed. 1 atm will not be the same amount of air. The box of steam is already full of steam. 1 kg of H2O is 1 kg. $\endgroup$ – paparazzo Mar 1 '17 at 10:08

13 Answers 13


First of all, it is impossible to have $1L$ of liquid water in vapor form in a $1L$ container. It is difficult for liquid form and the gaseous form to occupy the same volume. The gas molecules would be as close to each other as they were in the liquid form.

However, looking at your last paragraph, it can be inferred what you are actually asking for. I'll consider a very large box instead of a $1L$ box to answer this question.

What does a weighing scale measure?

A weighing scale measures the force applied on the weighing scale's platform by the test object.

Measuring the weight of liquid water

enter image description here

Consider a hypothetical situation where the water exists only in liquid form. There is no vapour pressure. Assume that the box is heavy enough that effects due to atmospheric pressure can be ignored.

The force which the box applies is given by:

$$F = m_{box}g + m_{water}g$$

The scale will read $(m_{box} + m_{water})g$.

Measuring the weight of gaseous water

enter image description here

The gas inside the container applies a pressure on the sides of the container.

Pressure on the top of the container = $P_{top}$

Pressure on the bottom of the container = $P_{top} + \rho gh$

The pressure on the top will help in reducing the force applied by the box on the measuring scale's platform and the pressure applied at the bottom will help to increase the weight applied by the box.

The horizontal forces applied on the sides of the box by the gas will cancel out neatly to give a net horizontal force of $0N$.

The force applied by the box on the scale is given by:

$$F = m_{box}g + (P_{bottom} - P_{top})A$$

$$F = m_{box}g + (P_{top} + \rho gh - P_{top})A$$

$$F = m_{box}g + \rho ghA$$

$$V = Ah$$

$$F = m_{box}g + (\rho V)g = m_{box}g + m_{gas}g$$

If you have the same number of molecules in the gaseous state as there are in the liquid state, $m_{water}$ obtained in the previous case is equal to the $m_{gas}$ obtained in this case.

The scale will read $(m_{box} + m_{water})g$.

If you strictly meant $1L$ of gas and $1L$ of water, then the water will definitely weigh more as it is denser but this question would be very silly if that was the case.

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    $\begingroup$ Buoyancy? and what's in the rest of the box that has the liquid water in it? $\endgroup$ – Rick Feb 28 '17 at 12:50
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    $\begingroup$ Buoyancy has nothing to do. @Rick There is nothing inside other than the water; the empty space is irrelevant to the question. The OP is trying to understand if gasses do weigh. There is no need to complicate the answer with unnecessary details. $\endgroup$ – Yashas Feb 28 '17 at 12:57
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    $\begingroup$ Of course that's what the OP is trying to understand, and that's exactly why bringing buoyancy into the answer is important. The reason people think that gases don't have weight is that that they are buoyant. Pointing out that someone's intuition is wrong isn't very helpful unless you show them why it's wrong. $\endgroup$ – Rick Feb 28 '17 at 13:44
  • $\begingroup$ Is having 1kg of water vapor in 1L of volume impossible under all conditions, or just at conventional temperatures and pressures? In other words can you achieve it in a high pressure/temperature situation, or would it end up on the wrong side of the critical point (where liquid vs gas ceases to be a meaningful differentiation). $\endgroup$ – Dan Is Fiddling By Firelight Feb 28 '17 at 14:53
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    $\begingroup$ More precisely, if you attempt to put the gas under enough pressure, above 22.064MPa, the "critical point" for water, the water no longer behaves like a gas. Above either the critical temperature or critical pressure, the distinction between the phases vanishes, and instead of having liquids and gasses with very different properties, you have one "supercritical fluid" state. Arguably the OP could be asking about 1L of liquid water or 1L of supercritical water, which could be accomplished... but 1L of gaseous water can't happen. $\endgroup$ – Cort Ammon Mar 1 '17 at 0:21

The force the box exerts on the scale will be given by the difference between the force that the gas does downwards on the bottom of the box and the force that it applies upwards to the top of the box.

Both forces can be written as pressure times area $F = P S$ where $S$ is the area of the top and bottom parts of the box. The difference in pressure is just given by the column of gas inside $P_{bottom}-P_{top}=\rho g h$, with $h$ the height of the box. Then:

\begin{equation} (\text{force on the scale})=F_{\text{bottom}}-F_{\text{top}}= (P_{\text{bottom}}-P_{\text{top}})S=\rho ghS=\rho g V=mg, \end{equation} is exactly the weight of the gas, where $V=hS$ is the volume and $m=\rho V$ is the mass.

Now, if the question is: What weighs more, a litre of liquid water or a litre of vapour? The answer is of course the liquid, because it has more density.

If instead we're asking if the same mass of water inside the box exerts less force of a scale, the answer is no, because the force it exerts is exactly its weight, as shown above.

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    $\begingroup$ Does this account for the density of air outside the box? given the box must be > 1L and the temperature is > 100°C isn't there buoyant force from air pressure ? $\endgroup$ – James Barrass Feb 28 '17 at 9:45
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    $\begingroup$ @YashasSamaga If we take into account buoyancy, 1kg of steam does exert less force on a scale than 1kg of liquid water. The weight of the box is irrelevant. $\endgroup$ – T. Verron Feb 28 '17 at 12:34
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    $\begingroup$ How? The gas is inside the box. Buoyancy is irrelevant here. The gas cannot affect its own enclosure. When I said the box must be heavy, I meant the weight must be large enough that the buoyant force due to atmospheric air won't have any effect on the box. $\endgroup$ – Yashas Feb 28 '17 at 12:36
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    $\begingroup$ @YashasSamaga The buoyant forces are dependent on the volume of the box, so the bigger box will have more buoyant lift force. Additionally, the OP specified weightless boxes, so the buoyant force will definitely dominate the weight of the box. In the case of the steam it will actually cause the box to float. $\endgroup$ – Rick Feb 28 '17 at 12:46
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    $\begingroup$ @Rick: From the question: " I acquired two weightless boxes of the same dimensions" (emphasis mine). So talking about the bigger box is meaningless. $\endgroup$ – Chris Feb 28 '17 at 12:47

Summarizing the question:

Start with two large weightless, identical boxes, and $2 \text{kg}$ of liquid water. Put $1 \text{kg}$ of water into each box. Heat one box such that none of the water molecules leave the box, but the water all boils into steam. Then compare the weights of the boxes when placed on a scale.

The weight of the boxes will indeed by the force pushing downward on the scale. This force will be the weight of the box (zero), plus the weight of the contents of the box, minus the buoyant force* of the air lifting up on the box. Since the boxes are identical, the buoyant force will be identical. And since the same mass of water went into both boxes initially the weight of the water in the boxes will also be identical.

However, the answer of which box will weigh more depends on what else is in each of the boxes:


If you have a vacuum filling the rest of the space, then there would be no additional mass or weight added. So the scales for both the liquid water box** and the gaseous water box would read $9.8 \text{N}$ minus the buoyant force*.

1 atm Air:

If you fill the remaining volume of the box with air at atmospheric pressure, then the amount of air that you would be adding to each box would be different. The liquid water will take up less space in the box, so more of the box will be filled with air. Thus more weight will be added to the box that has the liquid water in it and it will thus weigh more.

N moles of Air:

If you add the exact same number of of the same types of molecules to each box, then the masses would remain identical. However, the pressure in the box with the gaseous water would be much higher due to the higher temperature.

*Floating boxes:

Interestingly, gaseous water at the same temperature and pressure as gaseous air will be less dense, due to the lower molecular weight than nitrogen or oxygen. What that means is that the buoyant forces will actually be greater than the total weight of the boxes in three of the scenarios; Only the liquid water box filled the rest of the way with air would not just float away, making it very hard to place them on a scale.

It is likely this buoyancy of gases that gives you the intuition that an equal mass will weigh less as a gas than as a liquid.

**Phases and Vacuum:

In the case of the liquid water with a vacuum, it would instantly start boiling and depending on the temperature also possibly freezing due to the low pressure. However, as you watched the scale during these phase changes it would stay the same as the water boils and freezes, and then the frozen water eventually sublimates till it just has gaseous water in it like the other box.

  • $\begingroup$ While you are certainly correct that the water would be boiling if there is nothing above the water surface, this was not what the OP was looking for. The OP is having difficulty understanding if the gas has weight. The OP did mess up the question and multiple users have asked for clarification. From the last paragraph of the question, you can easily infer what the OP is actually looking for. $\endgroup$ – Yashas Feb 28 '17 at 13:38
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    $\begingroup$ @YashasSamaga thanks for pointing out that the boiling stuff really doesn't belong in the main body of the answer. I moved it to the end as a footnote. I still feel that comparing air to vacuum is really important to getting at the idea that gases have mass, as it's the everyday experience with buoyant air that gives people the intuition that gases don't have weight, and it's that intuition that we're trying to alter. So pointing out that it's our atmosphere that makes gasses seem weightless is an integral part of answering this question. $\endgroup$ – Rick Feb 28 '17 at 14:44
  • $\begingroup$ I like how you set this up because it leads to the question, "if I take a box with 1L of liquid water and apply heat until it boils, do I make it any lighter?" My intuition is that it wouldn't, but this is in conflict with the intuition that a box full of gas is lighter than a box full of liquid, which hopefully helps with understanding. $\endgroup$ – KC Baltz Mar 1 '17 at 22:02

This can be answered very simply: No. They will weigh exactly the same.

The two boxes are made of exactly the same matter occupying exactly the same total space. The only difference is in the internal arrangement of the components. That the gas exerts pressure doesn't mean anything -- internal forces cannot give a net force on the box for that would violate Newton's third law. This is then seen to make no difference at all as to the weight on a scale. Scale weight is two things (assuming no accelerating frames or other weirdness like that) -- the downward pull of gravity on the box plus the upward buoyant force on the box from the surrounding air. Both are exactly the same in each case by statement of the problem (same mass and same volume of air displaced by each container). So there is no difference in measured weight. We don't even need to do math -- a conceptual analysis is sufficient to answer the question.

EDIT: This is, of course, only considering classical mechanics alone. If our scale could be arbitrarily sensitive and we had a perfectly uniform 1-gee gravity field, then Einstein's relativity would come into play and it would say the steam box would weigh slightly more, due to the higher energy state it has versus the other box. The amount of mass (not weight-force) equals $\frac{n \Delta H_{\mathrm{vap}}}{c^2}$ where $n$ is the moles (not mass) of water and $\Delta H_{\mathrm{vap}}$ the heat of vaporization at the ambient temperature. For 1 kg of water this is about 0.03 micrograms. Local fluctuations in gravity would make this irrelevant to a weighing scale.


Skip the difference between mass and force. They both experience the same gravity.

Yes a liter of water is 1 kg

A liter of steam is about 1/1000 of that or about 1 g. More precisely 1.67 g (at 100C and 1 atm).

If you mean put 1 L of water in a 1 L box and heat it to 100C or even 200C then it will also be 1 kg. But it will NOT be steam. In a 1L box it does not have room to expand into a vapor. At 200C it will be about 15 atm of pressure. So have a strong weightless box.

I think you have butchered the question - I think you mean to ask.

The mass of 1 kg of water at 70 F and 1 atm and 1 kg of steam at 212 F and 1 atm is exactly the same.

The volume will be significantly different. About 1000 x different.

Weight is force and the weight will be different due to buoyancy. I don't have time to run the numbers right now.

density of water is 1 kg/L
density of steam is 0.0006 kg/ L

With buoyancy I think the steam would actually float away as it is less dense than air. But I will leave that for others.


Actually, if you consider relativistic effects, the box with steam will weight more, not less, since it has more energy and hence more mass. But ignoring these effects, both boxes should weight exactly the same: you have the same number of water molecules in each box, and gravity effect on each molecule is exactly the same.

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    $\begingroup$ If you're concerned about relativity, you should be concerned about the difference in gravitational strength varying with the height of box, relativity would increase the weight by about $10^{-12}$ while the vapor being higher in the box would (assuming a foot tall box) would decrease the weight by $10^{-8}$ both effects would be swamped by considering what else might be in the box. $\endgroup$ – Rick Feb 28 '17 at 16:40

If both boxes are the same size and weight, contain the same mass of water, and the same mass of air, the weight of both boxes will be the same, and the buoyant force on the boxes from the air that they displace will be the same. With "all things being equal", the two boxes will weigh the same when put on the electronic scale. However, all things are not equal. The box with the steam in it is substantially hotter than the box with the water in it. Due to this, the hot box will heat up the pan of the electronic scale, causing the air under the pan to heat up. This will produce a small amount of "lift", because the hot air is less dense than the air that is under the pan of the scale when the "water" box is weighed. Assuming a high precision for the hypothetical electronic scale, this will lead to a lower weight reading for the box containing steam than the box containing water.


Weight is proportional to mass, end of story, the two boxes will weigh the same. However to contain the mass of 1kg of steam in a ten litre box will require an enormous pressure. The volume of an ideal gas (which steam is not but neglect that for a moment) is 22.4 litres per mole at STP. One mole of water weighs 18g, so 1kg of steam will have a volume of 1244 litres at STP. At 100C the volume will be higher by a factor of around 373/298 in other words 1558 litres. So you want to put this in a 10L box which means applying a pressure of 156 atmospheres, which is a force of around 15.6 MPa, equivalent to 15.6 newtons per square millimetre. You will need oxygen cylinder type technology which will weigh a huge amount, which will be a tremendous burden on your accurate weighing scale.

Some other objections, a large pressure will elevate the boiling point of water, so you need to heat it up a lot more to ensure it remains gaseous at these huge pressures. Ideally you need to operate above the critical temperature of water which is 374C, resulting in a pressure ratio of 647/298 which is 270 atmospheres which is comfortably above the critical pressure.


Your question is poor. So poor that we can give an entire range of answers.

One one hand, the two contain the same rest mass but one has more kinetic energy. So its total mass has gone up by a non-measureable amount.

On another hand, the steam is probably higher in the box, and the force of gravity higher from the earth goes down. So its weight goes down by a non-measureable amount.

But you stated that the air was at 1 atm. Atm is a unit of pressure; boiling the water will increase the pressure through higher tempurature and through water atoms adding to the count of gas particles. So that implies your box had to leak a large amount of air to maintain that 1 atm, and thus the weight goes down by a measureable amount. In fact, it isn't possible to boil 1 kg of water in a 10 L box and maintain 1 atm pressure.

If we read your mind and ignore all of these problems, we get "a rigid box contains matter after a phase change, but the no matter entered or left, does is weight as measured by a practical scale change?" And the answer is no.

In general, when asking a physics question, it is best to describe a series of events as exactly as you can involving actions done by people or their proxies. For example:

You take two rigid, large boxes of 10 L each. At room tempurature and standard pressure, in each you place 1 kg of water. You then seal them.

Next, you heat one up past 100 C for long enough for the water within to boil. Nothing leaks out of either box.

You place them on a commercial scale with reasonable accuracy. Does one weigh more than the other?

The answer to this question is "no".


Your question is impossible (what now? Same amount of air plus 1 kg water vapor in the same volume and pressure as with 1 kg liquid water?). The question also appears to contain its own answer: "I acquired two boxes of the same dimensions and same weight. ... measure their respective weights ... what about the box containing steam?" Yes, what about it? It has the same weight, post ends here.

Ok, let's assume you meant something else — perhaps mass, not "weight"?

But whatever your setup, the answer is always very simple: The scale will measure the gravitational force the earth exerts on the gross mass of the box1 (i.e. packaging plus content, including all gas), minus the buoyant force of the box in the atmosphere. Packaging and buoyancy are identical2 for identical boxes, so any difference in "weight" (= force on the scale) must be from differences in the mass the boxes contain.

Now make up your mind: What's in the boxes (gases, fluids, solids)? Then you have the answer.

1 Or, if you want, the gravitational force that the gross mass of the box exerts on the earth. The attraction is mutual; it's just that the earth cares less.

2 Perhaps it's the buoyancy that confuses you. After all we seem to know that a volume containing as little as possible (hot air, helium, a vacuum) experiences the uplift we call buoyancy, but heavy objects like rocks or tanks do not; and the break-even point appears to be when the density of the contents is the density of air. But that's just our everyday approximation: In reality, all objects immersed in gas or fluid experience buoyancy; it's just that it is irrelevant to our everyday handling of heavy things. The amount of buoyancy exclusively depends on a body's volume (its displacement). It is the effect of pressure differentials between the underside and the upper side on the outside of the body; what's inside is irrelevant (and in the thought experiment of a perfect "black box" unknowable). If your boxes' contents was unknown, and they had the same mass (and mass distribution), they would behave exactly identically in all aspects. Whether the contents is a fluid or vapor or neutronium is unknowable and irrelevant.


It really all comes down to what is your definition of "weigh".

If you define weight in terms of the force exerted on the top of a weighing machine, then the weight of a helium-filled balloon is negative. The mass of that same balloon is positive. It's the fact that the mass of the balloon (skin and contents) is less than the mass of the air displaced, which causes the balloon to be bouyant, pulling upwards on your weighing machine rather than pushing down. In this case, the units of your result are Newtons (or kgf), not kg.

You'll get a similar result for steam (though you'd need a weighing machine in a room full of air hotter than 100C to stop the steam rapidly condensing). For a perfect gas at any particular temperature, a given volume contains a constant number of molecules. (The common formulation is that at standard temperature and pressure, one mole of gas occupies 22.4 litres). The molecular weight of water is around 16, that of air around 30.

But you said "box" not "balloon". If the box is a perfectly rigid pressure vessel, it won't expand and will prevent the water or steam from doing so. Therefore the force which the box exerts on the scales will not change. For any real pressure vessel, it will expand slightly and "weight" will change slightly. (Don't try this at home. The water won't become steam until the pressure in the box is very high, and live steam boiler explosions are deadly things).

Another definition of "weigh" means to determine the mass of an object by measuring the force that it exerts on a weighing machine and compensating for extraneous influences such as the mass of atmosphere displaced and local variations in gravity. Under this definition, the question is either tautologous, or is one about spotting sources of error in an actual "weighing" procedure, starting from the assumption that the mass of a quantity of water is not altered by boiling it. (well, not significantly. The added heat of sublimation has a tiny mass, as per Einstein's famous equation).

We should probably also note that "weigh" can commonly relate to determining a price for the substance on the scales, though that definition is unlikely to be an answer to this question.


The upward force is the same for both boxes (Archimedes), as is the downward force due to the mass of the boxes. But the second box doesn't contain the same amount of air: the situation in the second box is achieved by heating a box at room temperature, containing one litre of water and air at a lower pressure than 1 atmosphere (ending up at a pressure of 1 atmosphere at 100 degrees Celsius), which implies a lower mass of the contained air. If the air pressure at 100 degrees Celsius is 1 atmosphere, the air density is less.

The volume over which the air is distributed at 100 degrees Celsius increases by a factor that's less (from 9 litres to 10 litres) than the factor by wich the temperature increases (from 293 to 373 Kelvin).

Conclusion: The second box weighs indeed less.


The density of steam is $0.000804\frac{kg}{L}$.

Which is much less than of water which is $1\frac{kg}{L}$.

So $1L$ of gas would weigh $0.000804 kg$.

  • $\begingroup$ Sure, but I'm more interested in the WHY. A mathematical breakdown is definitely encouraged. $\endgroup$ – Nova Feb 28 '17 at 9:07
  • $\begingroup$ Because $ V \alpha T.$ As temperature of steam is more volume occupied by 1 kg water will be more in gas form than water form. And density=mass/volume. $\endgroup$ – ATHARVA Feb 28 '17 at 9:09
  • $\begingroup$ So the density of steam will naturally be less. $\endgroup$ – ATHARVA Feb 28 '17 at 9:10
  • $\begingroup$ The author was interested in knowing how the pressure of the gas would affect the weight. This answer does not discuss that matter. $\endgroup$ – Yashas Feb 28 '17 at 12:17
  • $\begingroup$ That is only true for standard temperature and pressure. He will have a pressurized box of steam (that is probably so hot its a plasma with the scales he wants :P). $\endgroup$ – Greg Mar 3 '17 at 3:56

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