Question

Figure shows a cylindrical tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by 1.5 nRT. The tube is divided into two equal parts by a fixed diathermic wall. Initially, the pressure and the temperature are p1, T1 on the left and p2, T2 on the right. The system is left for sufficient time so that the temperature becomes equal on the two sides. (a) How much work has been done by the gas on the left part? (b) Find the final pressures on the two sides. (c) Find the final equilibrium temperature. (d) How much heat has flown from the gas on the right to the gas on the left?
Figure

Answer 1

(a) Work has been done by the gas on the left part is  $T=\frac{\left(P_{1}+P_{2}\right) V}{2 n R}$

(b) Final pressures on the two sides $P_{2}^{\prime}=\frac{P_{2} T}{T_{2}}=\frac{P_{2}}{T_{2}} \frac{T_{1} T_{2}\left(P_{1}+P_{2}\right)}{P_{1} T_{2}+P_{2} T_{1}}$

(c) Final equilibrium temperature is $T=\frac{T_{1} T_{2}\left(P_{1}+P_{2}\right)}{P_{1} T_{2}+P_{2} T_{1}}$

(d) Heat has flown from the gas on the right to the gas on the left is $\frac{3 P_{1} P_{2} V}{4}\left(\frac{T_{2}-T_{1}}{P_{1} T_{2}+P_{2} T_{1}}\right)$

Explanation:

Let numbers of moles is $n_1,\, T_1 \,and \,n_2,\, T_2$

(a) So the final chamber volume is calculated to be the same as the diathermic wall.

ΔV = 0, hence ΔW= PΔV = 0

By state equation in chamber one and second

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

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

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

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

$n=n_{1}+n_{2}$

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

$n=\frac{V}{2 R}\left(\frac{P_{1}}{T_{1}}+\frac{P_{2}}{T_{2}}\right)$

$n=\frac{V}{2 R}\left(\frac{P_{1} T_{2}+P_{2} T_{1}}{T_{1} T_{2}}\right)$

$U=n C_{V} T$

$n C_{V} T=1.5 n R T$

$C_{V}=1.5 R$

The internal energy in the first and second chambers is supplied by

$U_{1}=n_{1} C_{V} T_{1}=n_{1} 1.5 R T_{1}$

$U_{2}=n_{2} C_{V} T_{2}=n_{2} 1.5 R T_{2}$

$U=U_{1}+U_2$

$1.5 n R T=n_{1} 1.5 R T_{1}+n_{2} 1.5 R T_{2}$

$1.5 n R T=1.5 R\left(n_{1} T_{1}+n_{2} T_{2}\right)$

$nT=(n_1 T_1+n_2 T_2)$

$n T=\frac{P_{1} V}{2 R T_{1}} T_{1}+\frac{P_{2} V}{2 R T_{2}} T_{2}$

$n T=\frac{\left(P_{1}+P_{2}\right)}{2 R} v$

$T=\frac{\left(P_{1}+P_{2}\right) V}{2 n R}$

$T=\frac{\left(P_{1}+P_{2}\right) V}{2 n R}$

$T=\frac{\left(P_{1}+P_{2}\right) V}{2 R \frac{V}{2 R}\left(\frac{P_{1} T_{2}+P_{2} T_{1}}{T_{1} T_{2}}\right)}=\frac{T_{1} T_{2}\left(P_{1}+P_{2}\right)}{P_{1} T_{2}+P_{2} T_{1}}$  eqn..(1)

(b) In the first and second compartments and  let final strain.

$\frac{P_{1} V}{\frac{2}{T_{1}}}=\frac{\frac{P_{1}^{\prime} V}{2}}{T}$

$P_{1}^{\prime}=\frac{P_{1} T}{T_{1}}$

From equation 1

$P_{1}^{\prime}=\frac{P_{1}}{T_{1}} \frac{T_{1} T_{2}\left(P_{1}+P_{2}\right)}{P_{1} T_{2}+P_{2} T_{1}}$

Likewise

$P_{2}^{\prime}=\frac{P_{2} T}{T_{2}}=\frac{P_{2}}{T_{2}} \frac{T_{1} T_{2}\left(P_{1}+P_{2}\right)}{P_{1} T_{2}+P_{2} T_{1}}$

(c) The final temperature is set to

$T=\frac{T_{1} T_{2}\left(P_{1}+P_{2}\right)}{P_{1} T_{2}+P_{2} T_{1}}$ by eqn (1)  

(d) Loss of heat by right chamber

$n_{2} C_{V} T_{2}-n_{2} C_{V} T=\frac{P_{2} V}{2 R T_{2}} 1.5 R T_{2}-\frac{P_{2} V}{2 R T_{2}} 1.5 R \frac{T_{1} T_{2}\left(P_{1}+P_{2}\right)}{P_{1} T_{2}+P_{2} T_{1}}$

$=\frac{3 P_{2} V}{4}-\frac{3 P_{2} V}{4} \frac{T_{1}\left(P_{1}+P_{2}\right)}{P_{1} T_{2}+P_{2} T_{1}}$

$=\frac{3 P_{2} V}{4}\left(1-\frac{T_{1}\left(P_{1}+P_{2}\right)}{P_{1} T_{2}+P_{2} T_{1}}\right)$

$=\frac{3 P_{2} V}{4}\left(\frac{P_{1} T_{2}+P_{2} T_{1}-T_{1}\left(P_{1}+P_{2}\right)}{P_{1} T_{2}+P_{2} T_{1}}\right)$

$=\frac{3 P_{1} P_{2} V}{4}\left(\frac{T_{2}-T_{1}}{P_{1} T_{2}+P_{2} T_{1}}\right)$