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%
Verified Correct Answer: C.
Physical Principles & Numerical Solution
\(\frac{P_2V_2}{T_2}\) = \(\frac{P_1V_1}{T_1}\)
P2 = 2P1
T1 = 2T2
\(\frac{2P_1V_2}{T_2}\) = \(\frac{P_1V_1}{2T_1}\)
Change in volume = \(V_2 - V_1\)
% change = \(\frac{V_1 - \frac{1}{4}V_1}{V_1}\) X 100%
= 75%
Source: Myschool Past Questions Archive (Record #28715)