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Thermodynamics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsThermodynamics3 Marks
Question
Calculate the fall in temperature when a gas initially at $72°C$ is expanded suddenly to eight times its original volume. Given y $=\frac53.(\therefore\text{V}_2=8\text{x}\text{ c.c.})$
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Answer

Let, $V_1 = x c.c.; T_1 = 273 + 72 = 345K; \gamma=\frac53; T_2 =$ ? Using the relation $\text{T}_1\text{V}^{\gamma-1}_1=\text{T}_2\text{V}^{\gamma-1}_2$
$\therefore\text{T}_2=\text{T}_1\Big(\frac{\text{V}_1}{\text{V}_2}\Big)^{\gamma-1}$ $=345\times\Big(\frac{\text{x}}{8\text{x}}\Big)^\frac23$
$=345\times\Big(\frac18\Big)^\frac23$
Taking $\log$ both sides, we get $\log\text{T}_2=\log\ 345-\frac23\log8$
$=2.5378-\frac23(0.9031)$
$=2.5378-0.6020$ $=1.9358$ $\text{T}_2=86.26\text{ K}$
$\therefore$ Fall in temperature = 345 - 86.26 = 258.74K.

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