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

A.

0%

B.

25%

C.

75%

D.

300%

E.

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%