Question

A non-conducting piston divides an adiabatic container into two equal parts. Such that piston is in equilibrium and temperature on both side is also same as $T_0$. Now $I^{st}$ chamber is heated such that piston moves very slowly until volumes of $I{I}^{nd}$ chamber is reduced to $\frac{1}{4}$ of the initial. In both chambers, same monatomic gas is filled such that the number of moles are same. The correct statement(s) is/are:
$(2^{4/3}=2.5)$

• A. Heat supplied to $1^{st}$ chamber is $27nR{T}_0$
• B. Work done by gas of $1^{st}$ chamber is $\frac{9}{4}nR{T}_0$
• C. Final temperature of $2^{nd}$ chamber is $\frac{5}{2}{T}_0$
• D. Final temperature of $1^{st}$ chamber is $\frac{35}{2}{T}_0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Heat supplied to $1^{st}$ chamber is $27nR{T}_0$
B Work done by gas of $1^{st}$ chamber is $\frac{9}{4}nR{T}_0$
C Final temperature of $2^{nd}$ chamber is $\frac{5}{2}{T}_0$
D Final temperature of $1^{st}$ chamber is $\frac{35}{2}{T}_0$

(take $2^{4/3}=2\cdot 5)$
W.D on piston = 0
$W_L+W_R=0$
for $I{I}^{nd}$ chamber Adiabatic process
$T_0{V}_0^{\frac{5}{3}-1}=T_f\cdot (\frac{V_0}{4}{)}^{\frac{5}{3}-1}$

$T_f=T_0{4}^{2/3}=T_0(2{)}^{4/3}=2\cdot 5T_0$

$W_R=-\Delta U=-n\frac{3}{2}R(2\cdot 5T_0-T_0)=-\frac{9}{4}nR{T}_0$

So. $W_L=\frac{9}{4}nR{T}_0{P}_0{V}_0\frac{5}{3}=P(\frac{V_0}{4}{)}^{\frac{5}{3}}$

$\Rightarrow P_f=P_0(4{)}^{\overline{3}}=P_0(4{)}^{\overline{3}}$

$\frac{P_0\times 4^{5/3}\frac{7V_0}{4}}{T_f}=\frac{P_0{V}_0}{T_0}{T}_f=7\cdot (4{)}^{\frac{2}{3}}{T}_0$

$T_L=\frac{35}{2}{T}_0$
Heat supplied

$Q=\frac{9}{4}nR{T}_0+n\frac{3}{2}R(\frac{35}{2}-1)T_0=(\frac{9}{4}+\frac{99}{4})nRT$

$=\frac{108}{4}nR{T}_0=27nR{T}_0$