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Q.

Thermodynamic process is shown below on a PV-diagram for one mole of an ideal gas. If $V_{2} = 2 V_{1}$ , then the ratio of $\frac{T_{2}}{T_{1}}$ is

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a

$\frac{1}{\sqrt{2}}$

b

$2$

c

$\frac{1}{2}$

d

$\sqrt{2}$

answer is C.

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Detailed Solution

$P V^{1 / 2} = c o n s \tan t$ $∴ T V^{r - 1} = K$ or $T α V^{1 - r}$

$\frac{T_{2}}{T_{1}} = {(\frac{V_{2}}{V_{1}})}^{(1 - r)}$

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