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$If the temperature of an ideal gas is doubled and the volume is halved, what happens to the pressure?$

A. $4 P$
B. $P$
C. $2 P$
D. $P /2$

Show correct answer & step-by-step solution

Correct answer: A — $4 P$

Solution

Using the ideal gas law P V = n R T, we have P = \frac{n R T}{V} . If T becomes 2 T and V becomes V /2, then the new pressure P^{'} = \frac{n R ( 2 T )}{V /2} = 4 \frac{n R T}{V} = 4 P .

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