Question 129 of 190

Thermodynamic process is shown below on a $P-V$ diagram for one mole of an ideal gas. If ${V}_{2}=2{V}_{1}$, then the ratio of temperature $\dfrac{{T}_{2}}{{T}_{1}}$ is :

A$\sqrt{2}$B$\dfrac{1}{\sqrt{2}}$C$\dfrac{1}{2}$D$2$

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Answer: (A) $\sqrt{2}$

$P{V}^{1/2}=c$
$\dfrac{nRT}{V}{V}^{1/2}=c$
$T={c}^{1}{V}^{1/2}$
$\dfrac{{T}_{2}}{{T}_{1}}={\left(\dfrac{{V}_{2}}{{V}_{1}}\right)}^{1/2}={\left(\dfrac{2{V}_{1}}{{V}_{1}}\right)}^{1/2}$
$\dfrac{{T}_{2}}{{T}_{1}}=\sqrt{2}$