How is an Isobaric process consistent with the first law of thermodynamics?

Question

Suppose there is an isobaric process whereby the volume of a gas decreases while the pressure of the gas remains constant. Work is done on the gas to compress it.

Since the change in internal energy = Q + W from the first law, there's a +W so the internal energy of the gas should increase.

However, this is inconsistent with Charles law whereby Volume of a gas is proportional to Temperature, since volume has decreased, temperature and hence also internal energy must have decreased.

What am I missing?

Answer 1

What you're missing is that the temperature must necessarily be lower after the gas is compressed and the pressure is held constant. While the work done on the gas is positive, the heat transferred away from the gas is even greater than the work performed. The isobaric specific heat of a gas is always greater than the gas constant.

Answer 2

Internal energy is directly proportional to the temperature of a given system. Consequently, a decrease in temperature (due to volume reduction in your case) will cause a net decrement in the internal energy.

Using the fact that the temperature decreases and the following deduction that the internal energy decreases too, you will come to an intuitive conclusion that a larger amount of heat will escape the system that will overweigh the external work resulting in a net negative change of internal energy.

Answer 3

How can you compress a gas and keep it at constant pressure?

If you compressed a gas adiabatically both the pressure and the temperature of the gas would rise.

So in both the examples that you have given, to keep the pressure of the gas constant whilst a gas is compressed, you must abstract heat from the gas and the temperature of the gas, and hence its internal energy, will decrease.
In a Charles's law apparatus to reduce the volume of the gas you would put the apparatus in water at a lower temperature.

When a gas is compressed at constant pressure:

• Heat gained by the gas, $Q = m\,C_{\rm P} \Delta T=m\frac 52 R \Delta T$.

• Work done by the gas, W = $P\Delta V = mR\Delta T$.

• Change of internal energy of gas, $\Delta U = \frac 32mR\Delta T$.

Noting that $\Delta T$ is negative, $\Delta U = Q - W \Rightarrow \frac 32mR\Delta T= \frac 52 m R \Delta T- mR\Delta T$