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?

30.0 J performed by the gas

4.00 J performed on the gas

0 J

30.0 J performed on the gas

Answer 1

Answer:

0 J

Explanation:

The work done by an ideal gas is given by the equation:

$W=p\Delta V$

where

W is the work done by the gas

p is the pressure of the gas

$\Delta V$ is the change in volume of the gas

In this problem, we have a gas kept at a constant volume of

V = 2.00 L

This is an isochoric process (constant volume). Since the volume is constant, the change in volume is zero:

$\Delta V=0$

And therefore, this means that the work done by the gas is zero:

W = 0