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First Law of Thermodynamics

A non conduct...

Question

A non conducting piston divides an adiabatic container into two equal parts. Such that piston is in equilibrium and temperature on both sides is also same as $T_{0}$ . Now the $1^{s t}$ chamber is heated such that piston moves very slowly until volume of $2^{n d}$ chamber is reduced to ${\frac{1}{4}}^{t h}$ of the initial. In both chambers, the same monatomic gas is filled such that the number of moles are same. $(2^{4 / 3} = 2.5)$

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Solution

W.D on piston = 0
$W_{L} + W_{R} = 0$
for $| |^{n d}$ chamber Adiabatic process
$T_{0} V_{0} \frac{5}{3} - 1 = T_{f} . (\frac{V_{0}}{4})^{\frac{5}{3} - 1}$
$T_{f} = T_{0} 4^{\frac{2}{3}} = T_{0} (2)^{\frac{2}{3}} = 2.5 T_{0}$
$W_{R} = Δ U = - n \frac{3}{2} R (2.5 T_{0} - T_{0}) = - \frac{9}{4} n R T_{0}$
So. $W_{L} = \frac{9}{4} n R T_{0} P_{0} V_{0}^{\frac{5}{3}} = P (\frac{V_{0}}{4})^{\frac{5}{3}}$
$\Rightarrow P_{f} = P_{0} (4)^{\frac{5}{3}} = P_{0} (4)^{\frac{5}{3}}$
$\frac{P_{0} 4^{\frac{5}{3}} \frac{7 V_{0}}{4}}{T_{f}} = \frac{P_{0} V_{0}}{T_{0}} T_{f} = 7. (4)^{\frac{2}{3}} T_{0}$
$T_{L} = \frac{35}{2} T_{0}$
Heat supllied
$Q = \frac{9}{4} n R T_{0} + n \frac{3}{2} R (\frac{35}{2} - 1) T_{0} = (\frac{9}{4} + \frac{99}{4}) n R T$
$= \frac{108}{4} n R T_{0} = 27 n R T_{0}$

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