Question

An ideal gas is kept at constant volume of 2.00 L as its temperature is increased, raising the pressure from 15.0 kPa to 30.0 kPa. What work is performed by or on the gas during this process?

A) 4.00 J performed on the gas

B) 0 J

C) 30.0 J performed on the gas

D) 30.0 J performed by the gas

Answer 1

The correct answer is: 
B) 0 J

Let's see why. The work done by the gas is equal to
$W=p \Delta V$
where 
p is the pressure
$\Delta V$ is the variation of volume of the gas.

However, the problem says that the volume is kept constant (2.0 L), so the variation of volume is zero: $\Delta V=0$, and the work done is zero as well.