Question

one mole of an ideal gas at temperature T1 expends according to the law $P/V^2=a\left(constant\right)$. the work done by the gas till the temperature of gas become T2 is:

• A. $\frac{1}{2}\times R(T_2-T_1)$
• B. $\frac{1}{3}\times R(T_2-T_1)$
• C. $\frac{1}{4}\times R(T_2-T_1)$
• D. $\frac{1}{5}\times R(T_2-T_1)$

Answer 1

The correct option is A $\frac{1}{2}\times R(T_2-T_1)$

$\text{Solution}:$

$W={\int }_{V_1}^{V_2}PdV={\int }_{V_1}^{V_2}KVdV$

$(\therefore \frac{p}{V}=constant)$

$\therefore W=\frac{1}{2}k(V_2^2-V_1^2)$

PV = RT

But p = KV

$\therefore K{V}^2=RT$

or $K(V_2^2-V_1^2)=R(T_2-T_1)$

$\therefore W=\frac{R}{2}(T_2-T_1)$

$\text{Hence A is the correct option}$