Q. The internal energy of a gas is given by $U=2 P V$. It expands from $V_{0}$ to $2 V_{0}$ against a constant pressure $P_{0}$. The heat absorbed by the gas in the process is :

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A

$2 P_{0} V_{0}$

27%

B

$4 P_{0} V_{0}$

44%

C

$3 P_{0} V_{0}$

12%

D

$P_{0} V_{0}$

17%

Solution:

$W =P_{0} \Delta V=P_{0} V_{0}$
and $\Delta U =2 P_{0}\left(2 V_{0}-V_{0}\right)=2 P_{0} V_{0}$
$\Rightarrow Q =W+\Delta U=3 P_{0} V_{0}$

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