Question:
Thermodynamic process is shown below on a $\mathrm{P}-\mathrm{V}$ diagram for one mole of an ideal gas. If $\mathrm{V}_{2}=2 \mathrm{~V}_{1}$ then the ratio of temperature $\mathrm{T}_{2} / \mathrm{T}_{1}$ is :
Correct Option: , 3
Solution:
$\mathrm{PV}^{1 / 2}=\mathrm{c}$
$\frac{\mathrm{nRT}}{\mathrm{V}} V^{1 / 2}=\mathrm{c}$
$T=c^{1} V^{1 / 2}$
$\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{1 / 2}=\left(\frac{2 \mathrm{~V}_{1}}{\mathrm{~V}_{1}}\right)^{1 / 2}$
$\frac{T_{2}}{T_{1}}=\sqrt{2}$
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