Question

A gas expands and does PV work on its surroundings equal to 319 J. At the same time, it absorbs 136 J of heat from the surroundings. Calculate the change in energy of the gas. Note: PV work means work done by a changing volume against constant pressure. Enter your answer in scientific notation.

Answer 1

Answer:

The change in energy of the gas during the process is $-1.83\times 10^{2}$ joules.

Explanation:

We can represent this process by the First Law of Thermodynamics, in which gas does work on its surroundings and absorbs heat from there to describe its change in energy. In other words:

$Q_{in} - W_{out} = \Delta E$

Where:

$Q_{in}$ - Heat absorbed by the gas, measured in joules.

$W_{out}$ - Work done by the gas, measured in joules.

$\Delta E$ - Change in energy, measured in joules.

If we know that $Q_{in} = 1.36\times 10^{2}\,J$ and $W_{out} = 3.19\times 10^{2}\,J$, the change in energy of the gas is:

$\Delta E = 1.36\times 10^{2}\,J-3.19\times 10^{2}\,J$

$\Delta E = -1.83\times 10^{2}\,J$

The change in energy of the gas during the process is $-1.83\times 10^{2}$ joules.