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Ideal Gas Equation

Three mole of...

Question

Three mole of an ideal gas at $300$ K are isothermally expanded to five times its volume and heated at this constant volume so that pressure is raised to its initial volume before expansion. In the whole process $83.14 k J$ heat is required.
If Poisson's ratio $(γ)$ for the gas is $x$ , then $\frac{100 x + 8}{30}$ is (nearest integer) :

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Solution

Apply Boyle's law for the isothermal expansion.
$\frac{P_{1}}{P_{2}} = \frac{V_{2}}{V_{1}} = \frac{5 V_{1}}{V_{1}} = 5$
At constant volume, apply gas law
$\frac{P_{2}}{T_{2}} = \frac{P_{3}}{T_{3}} = \frac{P_{1}}{T_{3}}$ (as $P_{3} = P_{1})$
$\frac{P_{2}}{P_{1}} = \frac{T_{2}}{T_{3}} = \frac{T_{1}}{T_{3}}$ (since, the first expansion was isothermal, $T_{1} = T_{2}$ )
$\frac{T_{3}}{T_{1}} = \frac{P_{1}}{P_{2}} = 5$
$T_{3} = 5 T_{1} = 5 \times 300 = 1500 K$
$d Q = \frac{μ R}{γ - 1} d T + μ R T_{1} l o g_{e} \frac{V_{2}}{V_{1}}$
Susbtitute values
$83.14 \times 10^{3} J = \frac{3 \times 8.3}{γ - 1} (1500 - 300) + 3 \times 8.3 \times 300 l o g_{e} 5$
$83140 = \frac{29880}{γ - 1} + 12027 [l o g_{e} 5 = 1.61]$
$γ = 1.42 = x$
$\frac{100 x + 5}{30} = 5$ .

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