Question

When a fixed amount of ideal gas goes through an isobaric expansion A) its internal (thermal) energy does not change.B) the gas does no work.C) no heat enters or leaves the gas. D) its temperature must increase.E) its pressure must increase.

Answer 1

Answer: its temperature must increase.

Explanation:

In an isobaric process the pressure remains constant, which means the initial pressure and the final pressure will be the same.

In addition, during this thermodynamic process, the volume of the ideal gas expands or contracts in such a way that the variation of pressure $\Delta P$ is neutralized.

Now, according to the First law of Thermodynamics that establishes the conservation of energy:

$\Delta U=\Delta Q-\Delta W$   (1)

Where:

$\Delta U$ is the internal energy

$\Delta Q$ is the heat transferred

$\Delta W$ is the work

Now, for an isobaric process:

$\Delta W=P\Delta V$    (2)

Where:

$P$ is the pressure (always positive)

$\Delta V$ is the volume variation of the gas

Here we have two possible results:

-If the gas expands (positive $\Delta V$), the work is positive.

-If the gas compresses (negative $\Delta V$), the work is negative.

In this case we are talking about the first result (work is positive).

Then, according to the above, equation (1) can be written as follows:

$\Delta U=\Delta Q - P\Delta V$   (3)

Clearing $\Delta Q$:

$\Delta Q=\Delta U+P \Delta V$    (4)

Then, for an ideal gas in an isobaric process, part of the heat ($Q$) added to the system will be used to do work (positive in this case) and the other part will increase the internal energy, hence the temperature will increase as well.