A given mass of an ideal gas occupies a volume V at a temperature T and under a pressure p. If the pressure is increased to 2p and the temperature reduced to $\frac{1}{2T}$ , then the percentage change in the volume of the gas is

25%

75%

300%

400%

Correct answer is C

$\frac{P_2V_2}{T_2}$ = $\frac{P_1V_1}{T_1}$

P₂ = 2P₁

T₁ = 2T₂

$\frac{2P_1V_2}{T_2}$ = $\frac{P_1V_1}{2T_1}$

V₂ --V₁

% change = $\frac{V_1 - \frac{1}{4}V_1}{V_1}$ X 100%

= 75%

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