Question

A gas is compressed from 600cm3 to 200cm3 at a constant pressure of 450kPa . At the same time, 100J of heat energy is transferred out of the gas.Part AWhat is the change in thermal energy of the gas during this process?

Answer 1

Answer: 80J

Explanation:

According to the first principle of thermodynamics:  

"Energy is not created, nor destroyed, but it is conserved."  

Then this priciple (also called Law) relates the work and the transferred heat exchanged in a system through the internal energy $U$, which is neither created nor destroyed, it is only transformed. So, in this especific case of the compressed gas:

$\Delta U=Q+W$  (1)

Where:

$\Delta U$ is the variation in the internal (thermal) energy of the system (the value we want to find)

$Q=-100J$ is the heat transferred out of the gas (that is why it is negative)

$W$ is the work is done on the gas (as the gas is compressed, the work done on the gas must be considered positive )

On the other hand, the work done on the gas is given by:

$W=-P \Delta V$  (2)

Where:

$P=450kPa=450(10)^{3}Pa$ is the constant pressure of the gas

$\Delta V=V_{f}-V_{i}$ is the variation in volume of the gas

In this case the initial volume is $V_{i}=600{cm}^{3}=600(10)^{-6}m^{3}$ and the final volume is $V_{f}=200{cm}^{3}=200(10)^{-6}m^{3}$.

This means:

$\Delta V=200(10)^{-6}m^{3}-600(10)^{-6}m^{3}=-400(10)^{-6}m^{3}$  (3)

Substituting (3) in (2):

$W=-450(10)^{3}Pa(-400(10)^{-6}m^{3})$  (4)

$W=180J$  (5)

Substituting (5) in (1):

$\Delta U=-100J+180J$  (6)

Finally:

$\Delta U=80J$  This is the change in thermal energy in the compression process.