Question

During an experiment, an ideal gas is found to obey an additional law $V{{P}^{2}}=$ constant. The gas is initially at a temperature T, and volume V. When it expands to a volume 2V, the temperature becomes:

• A. $\sqrt{2}T$
• B. T
• C. 4T
• D. 2T

Answer 1

Answer: (A)

$\sqrt{2}T$

Answer 2

For an ideal gas, the relation is $PV=RT$ ?(i) Given $V{{P}^{2}}=K$ ?(ii) Squaring equation (i) we get ${{P}^{2}}{{V}^{2}}={{R}^{2}}{{T}^{2}}$ ?(iii) Now dividing equation (ii) by (iii) $\frac{1}{V}=\frac{K}{{{R}^{2}}{{T}^{2}}}$ If volume V expands to volume 2V so ${{T}^{2}}\propto V$ Hence $\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{V}{2V}=\frac{1}{2}$ so, ${{T}_{2}}=\sqrt{2}\,{{T}_{1}}$