Question

A gas in a cylinder is held at a constant pressure of 1.80 * 105 Pa and is cooled and compressed from 1.70 m3 to 1.20 m3. The internal energy of the gas decreases by 1.40 * 105 J.
(a) Find the work done by the gas.
(b) Find the absolute value |Q| of the heat flow into or out of the gas, and state the direction of the heat flow.
(c) Does it matter whether the gas is ideal? Why or why not?

Answer 1

Answer:

The work done is $W=-0.9*10^5J$ and it is negative because the exterior is doing work on the gas.

The value of the heat flow is $Q=2.3*10^5J$ and is directed out of the gas, as this is being cooled and compressed.

Explanation:

The work done by the gas can be written as

$W=\int\limits^ {V_{c}}_{V_{h}}{P}\, dV$

and all the data is given in the problem, so

$W=P(V_{c}-V_{h})=-0.9*10^5J$

is the work necessary to compress the gas in the cylinder, wich is negative because the exterior is doing work on the gas.

Now, from first law of thermodynamics, we have that

$\Delta U=Q+W$

where $\Delta U$ is the internal energy difference, $W$ is the calculated work, and $Q$ is the heat flow that we wish to know, then

$Q=\Delta U-W=1.4*10^5J+0.9*10^5J=2.3*10^5J$

And the direction of the heat flow is outside of the gas, as it is being cooled and compressed.

Finally, it doesn't matter whether the gas is ideal or not, because we never use the Ideal Gas hypotesis, that $PV=nRT$, and $P$ is constant.