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From the ideal gas equation: $PV = NkT$, since Pressure times Volume = Energy, my understanding is that the total (internal) energy of $N$ molecules of a gas $= NkT$.

However from the kinetic theory equation: Average kinetic energy per molecule of gas = $\frac{3}{2} kT$ and hence the total kinetic energy for $N$ molecules = $\frac{3}{2} NkT$.

Since potential energy is considered non-existent in ideal gases, kinetic energy = internal energy. However the 2 formulae lead to different results. What is the reason for this, what am I missing?

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    $\begingroup$ Pressure x volume = energy is not a hard and fast rule. In this case, its only correct up to a factor of 2/3. $\endgroup$
    – jacob1729
    Commented May 9, 2019 at 15:51
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    $\begingroup$ Also note that rotational nor vibrational degrees of freedom, contributing to the gas molar heat capacity, do not contribute in pressure not mechanical work via p and V. $\endgroup$
    – Poutnik
    Commented May 9, 2019 at 15:55
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    $\begingroup$ Who says that the internal energy of an ideal gas (or its kinetic energy) is supposed to be PV? $\endgroup$ Commented May 9, 2019 at 17:05
  • $\begingroup$ If its not then what does the PV represent? $\endgroup$
    – IK-_-IK
    Commented May 9, 2019 at 18:41
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    $\begingroup$ It merely represents part of the equation of state PV=nRT, unrelated to the internal energy. Why to you think it is related to internal energy...simply because it has the same units?? $\endgroup$ Commented May 9, 2019 at 19:25

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Actually its incorrect to think that $$PV=energy,$$ just because they have same units.

Strictly speacking, its defined that $$dW=PdV$$ where $dW$ is work done in change in volume of $dV$.

So your upper derivation of kinetic energy from ideal gas equation is incorrect. The kinetic energy will be found using, equipartion law and degree of freedom.

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Since potential energy is considered non-existent in ideal gases, kinetic energy = internal energy. However the 2 formulae lead to different results. What is the reason for this, what am I missing?

While your first statement is true, please note that you're wrong in saying $PV = U$.

As I show below the correct relationship betweeen $PV$ and $U$ comes with a factor of $2/3$.

We first note that internal energy ($U$) of an ideal gas is only kinetic energy ($K$), as there is no potential of interaction. Therefore, firstly we have, $$ U = K $$ Now let us see from kinetic theory of gases, how we can calculate pressure ($P$) from kinetic energy ($K$). Physically, pressure is the force per unit area exerted by the gas atoms/molecules on the walls of the container.

Consider a gas of a large number $N$ of molecules, each of mass $m$, enclosed in a cube of volume $V = L^3$. When a gas molecule collides with the wall of the container perpendicular to the $x$ axis and bounces off in the opposite direction with the same speed (an elastic collision), the change in momentum is given by: $$ \Delta p = p_i - p_f = p_i - (-p_i) = 2mv_x $$ where $p_i$ represents the initial momentum of the molecule and $p_f$ represents the final momentum after the collision. Here $m$ is mass and $v_x$ the velocity in $x$-direction.

Let us say that the particle impacts one specific side wall only once during the time interval $\Delta t$, therefore, $$ \Delta t=\frac{2L}{v_x} $$ Therefore, the force exerted on the wall due to $N$ particles is, $$ F = N\frac{\Delta p}{\Delta t} = N\frac{m v_x^2}{L} $$ Assuming the motion of particles is isotropic, we get, mean squared speed $\left< v^2 \right> = \left< v_x^2 \right>/3$, and hence the force can be written as, $$ F = \frac{Nm \left< v^2 \right>}{3 L} $$ which leads us to the expression for pressure, force on an area $L^2$ $$ P = \frac{F}{L^2} = N\frac{m\left< v^2\right>}{3L^3}=\frac{2}{3}N\left(\frac{m\left< v^2 \right>}{2}\right) \frac{1}{V} = \frac{2}{3} \frac{K}{V} = \frac{2}{3} \frac{U}{V} \\ \implies PV = \frac{2}{3} U $$

Now, if you subsititute the values for average kinetic energy per molecule, you'll get the ideal gas equation.

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$PV~=~Nk_BT$

$\Rightarrow \frac{PV}{NT}~=~k_B$

The Ideal Gas Equation just says that the product of Pressure and Volume divided by the product of the Number of particles and Temperature is always constant and equal to the Boltzmann Constant. It doesn't tell anything about energy.

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