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1st Law of Thermodynamics

A cylindrical...

Question

A cylindrical container having non conducting walls is partitioned into two equal parts such that the volume of each part is $V_{0}$ . A movable non conducting piston is kept between the two parts. Gas on the left is slowly heated so that the gas on right is compressed up to volume $\frac{V_{0}}{8}$ . Find pressure and temperature on both sides if the initial pressure and temperature were $P_{0}$ and $T_{0}$ respectively. Also, find heat given by the heater to the gas. (number of moles in each part is n)

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Solution

Since the process on right is adiabatic therefore
$P V^{γ} = c o n s t a n t$
$\Rightarrow P_{0} V_{0}^{γ} = P_{f i n a l} (V_{0} / 8)^{γ} \Rightarrow P_{f i n a l} = 32 P_{0}$
$T_{0} V_{0}^{γ - 1} = T_{f i n a l} (V_{0} / 8)^{γ - 1} \Rightarrow T_{f i n a l} = 4 T_{0}$
Let volume of the left part is $V_{1}$
$\Rightarrow 2 V_{0} = V_{1} + \frac{V_{0}}{8} \Rightarrow V_{1} = \frac{15 V_{0}}{8}$ .
Since the number of moles on the left part remains constant therefore for the left part $\frac{P V}{T} = c o n s t a n t$ .
Final pressure on both sides will be same
$\Rightarrow \frac{P_{0} V_{0}}{T_{0}} = \frac{P_{f i n a l} V_{1}}{T_{f i n a l}} \Rightarrow T_{f i n a l} = 60 T_{0}$
$Δ Q = Δ U + W$
$Δ Q = n \frac{52}{2} (60 T_{0} - T_{0}) + n \frac{3 R}{2} (4 T_{0} - T_{0}) \Rightarrow Δ Q = \frac{5 n R}{2} \times 59 T_{0} + \frac{3 n R}{2} \times 3 T_{0} = 152 n R T$

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