Question

An adiabatic vessel of total volume V is divided into two equal parts by a conducting separator. The separator is fixed in this position. The part on the left contains one mole of an ideal gas (U = 1.5 nRT) and the part on the right contains two moles of the same gas. Initially, the pressure on each side is p. The system is left for sufficient time so that a steady state is reached. Find (a) the work done by the gas in the left part during the process, (b) the temperature on the two sides in the beginning, (c) the final common temperature reached by the gases, (d) the heat given to the gas in the right part and (e) the increase in the internal energy of the gas in the left part.

Answer 1

a) As the conducting wall has been set, The gas work done on the left part of the process is null Because the volume change will be zero due to the wall's fixed position.

(b) the temperature on the two sides in the beginning $T_{1}=\frac{P V}{4(\text { moles }) R}$.

(c) the final common temperature reached by the gases $T=\frac{P V}{3 R}$.

(d) the heat given to the gas in the right part and $\bmod e l s \frac{3 \times P V}{4 \times 3 \text { mole }}=P V$.

 (e) the increase in the internal energy of the gas in the left part $\mathrm{dU}=-\mathrm{d} \mathrm{Q}=-\frac{\mathrm{PV}}{4}$.

Explanation:

a) As the conducting wall has been set, The gas work done on the left part of the process is null Because the volume change will be zero due to the wall's fixed position.

(b) To the left:

Let the initial pressure be p at both sides of the wall.

$\text { Volume }=\frac{V}{2}$

n  = 1 (mole)

$T_1$ is the initial temperature  

Step 1:

using the gas of ideal equation

$\frac{P V}{2}=n R T_{1}$

$\frac{P V}{2}=R T$

$T_{1}=\frac{P V}{2(\text {moles}) R}$

Step 2:

To the right side  

n = 2 (moles)

$T_2$ is final temperature  

$\text { Volume }=\frac{V}{2}$

$\frac{P V}{2}=n R T_{2}$

$T_{1}=\frac{P V}{4(\text { moles }) R}$

c) The final common temperature reached by the gases,

U = 1.5nRT

T = Equilibrium temperature

$P_1$  =P$_2$= P  

$n_1$  = 1 mol  

$n_2$  = 2 mol  

Step 3:

Let $T_1$ and $T_2$ be initial Temperatures of the chamber  left and right.

For left side chamber

$\frac{P V}{2}=n_{1} R T_{1}$

$\frac{P V}{2}=R T_{1}$

$P V=2 R T_{1}$

$\frac{P V}{2 R}=T_{1}$

Step 4:

To the right chamber

$\frac{P V}{2}=n_{2} R T_{2}$

$\begin{aligned}&\frac{P V}{2}=2 R T_{2}\\&P V=4 R T_{2}\\&\frac{P V}{2 R}=T_{2}\end{aligned}$

Step 5:

$\begin{aligned}&n=n_{1}+n_{2}=3\\&U_{1}=n_{1} C_{V} T_{1}=C_{V} T_{1}\end{aligned}$

$\begin{array}{c}{U_{2}=n_{2} C_{V} T_{2}=2 C_{V} T_{2}} \\{U=U_{1}+U_{2}}\end{array}$

$\begin{array}{c}{3 C_{V} T=C_{V} T_{1}+2 C_{V} T_{2}} \\{3 T=T_{1}+2 T_{2}}\end{array}$

$3 T=\frac{P V}{2 R}+2 \frac{P V}{4 R}$

$3 T=\frac{P V}{2 R}+\frac{P V}{2 R}$

$3 T=\frac{P V+P V}{2 R}$

$3 T=\frac{2 P V}{2 R}$

$3 T=\frac{P V}{R}$

$T=\frac{P V}{3 R}$

(d) For right hand side

$\begin{array}{l}{\Delta Q=\Delta U \text { as } \Delta W=0} \\{\Delta U=1.5 n_{2} R(T-\mathrm{T} 2)}\end{array}$

If T is the final temperature and $T_2$ is the initial side 1 temperature, then we get

$\begin{aligned}&=1.5 \times 2 \times \mathrm{R}(\mathrm{T}-\mathrm{T} 2)\\&=1.5 \times 2 \times \frac{4 P V-3 P V}{4 \times 3 m o l e}\end{aligned}$

$\bmod e l s \frac{3 \times P V}{4 \times 3 \text { mole }}=P V$

(e) If dW = 0, then we get to use the first rule

$\begin{array}{l}{\mathrm{dQ}=-\mathrm{dU}} \\{\mathrm{dU}=-\mathrm{dQ}=-\frac{\mathrm{PV}}{4}}\end{array}$