# Tag Info

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The piston moves to left if $P_2\gt P_1$ and to right if $P_2\lt P_1$. In above answers, Lucas's beginning part is fine. $$P_1V_1=m_1RT_1$$ $$P_2V_2=m_2RT_2$$ with $P_1=P_2$,$V_1=V_2$,$T_2=T_1-ΔT$ We then have, $$\frac{m_2}{m_1}=\frac{T_1}{T1-ΔT}$$ When temperature on both side is raised by $\delta T$ $$P′_1V′_1=m_1RT′_1$$ $$P′_2V′_2=m_2RT′_2$$ with ...

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Rigorous calculation The first hypothesis is hand-wavy Aristotelian logic. Whatever does happen has to be explainable by the Ideal Gas Law. At all times, $p_1 = p_2$ must hold, otherwise, the piston would move to equalize the pressures. So, $$\begin{eqnarray} p_1 =& p_2 \\ \frac{N_1 k_b T_1}{V_1} =& \frac{N_2 k_b T_2}{V_2} \\ \frac{V_2}{V_1} ... 2 Let T_{20} be the initial temperature of tank 2 and T_{10}=T_{20}+\Delta T be the initial temperature in tank 1. Let \delta T be the equal rise in the temperature of both thanks. Assuming that the piston does not move, we would have$$p_{2f}=p_2\frac{T_{20}+\delta T}{T_{20}}$$and$$p_{1f}=p_1\frac{T_{20}+\Delta T+\delta T}{T_{20}+\Delta T}$$Since ... 1$$P_1V_1=m_1RT_1P_2V_2=m_2RT_2$$P_1=P_2,V_1=V_2,T_1=T_2+\Delta T Then, we have:$$\frac{m_2}{m_1}=\frac{T_1}{T_1-\Delta T}$$After heating:$$P'_1V'_1=m_1RT'_1P'_2V'_2=m_2RT'_2$$P'_1=P'_2,T'_1=T'_2+\Delta T Then, we have:$$V'_2=V'_1\left(\frac{m_2(T'_1-\Delta T)}{m_1T'_1}\right)=V'_1\left(\frac{T_1}{T_1-\Delta ...

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Your second thesis is right. At least the result. The explanation not, as the example in the answer of Diracology might illustrate. Assume, that the piston does not move (we fix it). So $V$ didn't change on either side. Obviuosly, $n$ didn't change on either side. Therefore, $p$ is proportional to $T$ (in each side separately!). So if $T$ rises by the same ...

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I actually did some of the algebra work and using the equations given, I managed to show that the pressure from 1 is less than 2. The volume being held constant is important here. Just to note I'm purely using algebra here, so please check through my working and see if it makes sense to you. For notation sake, I've set anything with b as before the tube is ...

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The piston cannot move if the pressures are maintained equal (assuming same cross section are at both extremes). Otherwise you would violate the second Newton's law. The fact that the second container has more energy does not imply the piston should move. Imagine two containers filled with the same gas linked by a piston. Assume same pressure and same ...

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I think the idea is basic. In the experiment you say that "The Volume of the containers also stays constant." Which the tubes in the diagram are part of the containers, the piston can not move in order to keep the volume constant. Now if the piston could move, then the volume would change, considering the Ideal Gas Law, the piston would move to the left and ...

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Let's start with the physical interpretation. We are considering an ideal gas of particles in equilibrium at some temperature $T$. Let's ask the following question: if the system is in equilibrium, why don't all particles have the same speed? Answer: because the particles interact through collisions. Imagine that one could prepare a system in such a way that ...

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The Maxwell-Boltzmann Distribution is a probability distribution - eg the distribution of speeds of particles in a gas. The area under the curve represents probability, not energy, so the total area under the curve must be 1, regardless of temperature. The y axis represents a probability density function, eg probability per unit interval of speed, while the ...

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After further research, I've concluded that the ideal gas law would work for Venus' supercritical fluid atmosphere, at least reasonably well enough for my curiosity, and certainly as well as it would here on Earth. My research took me to learning compressibility factor, equations of state, and several other real gas topics. From what I gather, a gas ...

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The ideal gas law assumes that there are no forces between molecules of a gas, and that the size of the molecules is negligible compared to the volume of the gas. When a gas becomes a liquid, these assumptions are clearly violated; and when it becomes supercritical, the density is typically still such that the second assumption is almost certainly not valid ...

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its really easy PV=NRT P(V+dV)=NR(T+dT) PV+PdV=NRT+NRdT PdV=NRdT=work done (w) you know N,R dT calculate w...

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Simplify the equation to make it more convenient to you by using the definition of a mole. A mole , generally denoted by 'n' is the mass of the substance taken divided my its molecular weight. On solving the only unknown in this equation , you get the mass of air contained in the volume you obtained. Now , you can just plug in this value into the definition ...

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The ideal gas law says you something about $p$, $V$ and $T$ in terms of the number of particles. That's nice, because it holds approximately for all gases. For a given $p$ and $T$ you know the number per volume - it is independent of the type of gas! Then why is the density of gases different? - because the mass per particle is different. So now you have to ...

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Consider an adiabatic container with a partition. Let one side be filled with an ideal gas and the other side is filled with vacuum.Now, keeping the pressure on this system constant , I remove the partition thereby changing the volume which in turn changes the temperature of the system and the pressure.Using the adiabatic relationships between P,V,T, the ...

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You can probably take advantage of measuring pressure at two different heights. Air pressure is caused by its weight, similar to liquid pressure. Therefore, it can be shown that pressure difference of two points can be related to air density: $$\Delta p=\rho g \Delta h$$ This equation gives $\rho$ (average mass density) as a function of $g$ (gravity ...

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Okay - probably you know, that weather physics is serious business and having even lot of measurements in many points you still can't calculate long enough evolution of a system to get results for e.g. weather in the next month. But you are asking for something else- just equation of gas density in open system depending on local temperature and pressure. ...

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To decrease the volume of the gas and hence increase its pressure work needs to be done. That is an external force must move through a distance. The work done can be found by working out the integral $\int P \;dV$ which is the area under a Pressure, $P$, against volume, $V$, graph.

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Pressure P is a compressive stress, and tension is related to tensile stress. So for the gas, the reversible work done by the surroundings on the gas is -PdV. And, for the elastic string, the reversible work done by the surroundings on the string is TdL. This last equation is the same as force times displacement. For the gas, it can be worked into ...

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Viscous dissipation of mechanical energy would be the dominant factor in characterizing the irreversibility. This would be readily captured by the tensorial form of the stress-(deformation rate) equation for a Newtonian (viscous) fluid. This describes the deformational behavior of a fluid in both simple shear flow as well as in much more complicated three ...

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For an ideal balloon-pop (no interactions with the walls of the balloon) $dU = 0$. In the case of an ideal gas, $U = U(T) \propto NkT$. So the temperature, strictly speaking, does not change when the balloon expands. You end up with a gas at the same temperature but with a lower density.

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turns out the answer was quite simple, just use Dalton's law and Amagat's law in combination with the ideal gas law. $$P=\sum_{i=1}^kP_i=P_1+P_2$$ $$V=\sum_{i=1}^kV_i=V_1+V_2$$ $$PV=n\overline R T$$ $$V_i=m_iv_i=\frac{m_i}{\rho_i}=\frac{n_iM_i}{\rho_i}$$ $$P=P_1+P_2=\frac{n_1\overline R T}{V}+\frac{n_2\overline R T}{V}=(n_1+n_2)\frac{\overline R T}{V}$$ ...

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You haven't given enough information to make such a determination. It is pointless to try to guess what tidbit you want to add or reveal next. The basic equation is PV=nRT - the ideal gas law. So $$(P_1+P_2)V = (n_1+n_2)RT$$ Edit - The gist here is that you have one volume. The gases mix in it. You can calculate the "fraction of the volume" that is one or ...

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The condition that the internal energy of an ideal gas is a function only of temperature follows empirically from the observed behavior of real gases in the limit of low pressures and high specific volumes. (It may also follow from statistical thermodynamics). In addition, the heat capacities of most real gases are observed to be a function of temperature ...

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I think that there is a confusion because there are two forms of the ideal gas equation. In terms of moles the ideal gas equation is $PV = nRT$ where $R$ is the molar gas constant and is a constant independent of the gas or gases which are being considered. In this equation there is no mention of the mass of the gas or the composition of the gas other ...

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An ideal gas is one that follows any of a number of linear relations such as $PV = k$ or $PV = nRT$. Many gases near room temperature and atmospheric pressure are nearly ideal and show this relationship pretty closely. All gasses deviate from the ideal as pressure increases and temperature decreases. As an example, if you cool the gas so much that it ...

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First of all, the term $PV=mv_x^2$ you got is due only to one molecule. You have to sum over all molecules, $$PV=m\sum v_x^2=N\frac{\sum v_x^2}{N}=Nm\bar v_x^2.$$ Assuming space isotropy you can write $$\bar v_x^2=\bar v_y^2=\bar v_z^2=\frac 13 \bar v^2.$$ Hence $$PV=\frac N3m\bar v^2.$$ Now you have to invoke the Equipartition Theorem It basically says ...

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Just a sketch. The kinetic theory shows that $$U=\frac{3}{2}PV$$ Hence $$dU=\frac{3}{2}PdV+\frac{3}{2}VdP$$ By 1st law, $$\delta Q=dU+PdV=\frac{5}{2}PdV+\frac{3}{2}VdP$$ Consider a small rectangle $R$ on the PV diagram, it can be shown that $$\oint_R \delta Q \ne 0$$ But $$\oint_R \frac{\delta Q}{PV}=0$$ In other words, $\frac{1}{PV}$ is the ...

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The net momentum of the particles in your volume $dA\; vdt$ is zero so on average only half of those particles will hit your wall and rebound. That is were your missing $\frac 12$ comes from.

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The derivation you quoted above is from the kinetic theory of gases. I think you need to revise the step in which you defined the force. See this Link: http://quantumfreak.com/derivation-of-pvnrt-the-equation-of-ideal-gas/ There is a much simpler way to derive the ideal gas equation: The three historically important gas laws derived relationships between ...

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Your approach is along the right lines and you need to make use that $dp = \rho(z) \, g \, dz$ and use the relationship between pressure, volume and temperature for n moles of gas as $P V = n R T$, so that $P = \frac{n}{V} R T = \rho R T$, with $\rho$ the density. If you assume that the temperature of the atmosphere is uniform then the pressure varies as ...

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