thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas.

Question:

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 $T_{2} / T_{1}$ is:

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

2. $\frac{1}{2}$

3. 2

4. $\sqrt{2}$

Correct Option: , 4

Solution:

From $\mathrm{p}-\mathrm{v}$ diagram,

Given $\mathrm{Pv}^{1 / 2}=$ constant $\ldots$ (i)

We know that

$\mathrm{PV}=\mathrm{nRT}$

$P \propto\left(\frac{T}{v}\right)$

Put in equation (i)

$\left(\frac{T}{v}\right)(v)^{1 / 2}=$ constant

$T \propto v^{1 / 2}$

$\frac{T_{2}}{T_{1}}=\sqrt{\frac{V_{2}}{V_{1}}}$

$\frac{T_{2}}{T_{1}}=\sqrt{\frac{2 \sigma_{1}}{V_{2}}}$

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