Heat and Thermodynamics 6 Question 38
43. The rectangular box shown in figure has a partition which can slide without friction along the length of the box.
Initially each of the two chambers of the box has one mole of a monoatomic ideal gas $(\gamma=5 / 3)$ at a pressure $p _0$, volume $V _0$ and temperature $T _0$. The chamber on the left is slowly heated by an electric heater. The walls of the box and the partition are thermally insulated. Heat loss through the lead wires of the heater is negligible. The gas in the left chamber expands pushing the partition until the final pressure in both chambers becomes $243 p _0 / 32$. Determine (a) the final temperature of the gas in each chamber and (b) the work done by the gas in the right chamber.
(1984, 8M)
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Answer:
Correct Answer: 43. (a) $T _1=12.94 T _0, T _2=2.25 T _0$ (b) $-1.875 R T _0$
Solution:
- (a) Applying $\frac{p V}{T}=$ constant for both the chambers.
$$ \begin{gathered} \frac{p _0 V _0}{T _0}=\frac{p V _1}{T _1}=\frac{p V _2}{T _2} \quad\left(\text { Here, } p=\frac{243 p _0}{32}\right) \\ \therefore V _1=\frac{32}{243}\left(\frac{T _1}{T _0}\right) V _0 \text { and } V _2=\left(\frac{32}{243}\right)\left(\frac{T _2}{T _0}\right) V _0 \end{gathered} $$
Further, $\quad V _1+V _2=2 V _0$
or $\left(\frac{16}{243}\right)\left(T _1+T _2\right)=T _0$ or $T _1+T _2=\frac{243}{16} T _0$
Applying the $p-T$ equation of adiabatic process to the right chamber,
$$ \left(\frac{T _0}{T _2}\right)^{5 / 3}=\left(\frac{243 p _0}{32 p _0}\right)^{1-5 / 3} $$
Solving this equation, we get
$$ \begin{array}{ll} & T _2=2.25 T _0 \\ \text { From Eq. (i), }, & T _1=12.94 T _0 \end{array} $$
(b) Work done by the gas in right chamber
$$ (\Delta Q=0, \text { adiabatic process) } $$
$$ \Delta W=-\Delta U=n C _V\left(T _i-T _f\right) $$
$$ =(1)\left(\frac{3}{2} R\right)\left(T _0-2.25 T _0\right)=-1.875 R T _0 $$
