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Thermodynamics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsThermodynamics2 Marks
Question
Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in Fig.

Find the work done when the gas is taken from state 1 to state 2.
✓

Answer

$\therefore\text{PV}^\frac{1}{2}= $ constant = K (given) or Work done for process from 1 to 2
$\text{WD}=\int\limits^{\text{v}_{2}}_{\text{v}_{1}}\text{P}.\text{dV}=\int\limits^{\text{v}_{2}}_{\text{v}_{1}}\frac{\text{K}}{\text{V}^\frac{1}{2}}\text{dV}=\text{K}\int\limits^{\text{v}_{2}}_{\text{v}_{1}}\text{V}^{-\big(\frac{1}{2}\big)\text{dV}}$$$
$\text{WD}=\text{K}\Bigg[\frac{\text{V}^\frac{1}{2}}{\frac{1}{2}}\Bigg]^{\text{v}_2}_{\text{v}_1}=2\text{K}\big[\sqrt{\text{V}_{2}}-\sqrt{\text{V}_1}\big]$
WD form $\text{V}_1\text{to}\text{V}_2,\text{i.e},\text{dW}=2\text{P}_1\text{V}_1^\frac{1}{2}\big[\sqrt{\text{V}_2}-\sqrt{\text{V}_1}\big]$
$=2\text{P}_2\text{V}_2^\frac{1}{2}\big[\sqrt{\text{V}_2}-\sqrt{\text{V}_1}\big]$

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