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: Option C
Explanation
(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%