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)

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, $T_2=\sqrt{2}{T}_1$