Question

An ideal gas expands while the pressure is kept constant. During this process, does heat flow into the gas or out of the gas? Justify your answer.

Short answer

An ideal gas expands while the pressure is kept constant because heat flows into the gas.

Step-by-step solution

Work done and laws of thermodynamics.

The work done in a volume change at constant pressures

$$\begin{gathered} F=pA \\ W=pA∆x=p∆V \\ W=p\left(V_2-V_1\right)=p∆V \end{gathered}$$

Where p =pressure of the gas,$V_2$ = final volume,$V_1=$ initial volume

Ideal gas equation:

$PV=nRT$

P= the pressure of the gas, V = the volume, n number of moles,

R= 8.314472J/mol.K = ideal gas constant, T temperature of the gas.

The total heat added to a gas in a constant-pressure process:

$\underset{~}{O}=W+∆U$

Q =the total heat added to the gas, W the work done by the gas on the surrounding,

$∆U=$the change in the internal energy which depends only on the temperature of the gas.

Direction of heat flow.

$$\begin{gathered} W=p∆V \\ ∆V>0=>Thegsexpands \\ W>0 \end{gathered}$$

$$\begin{gathered} pV=nRT \\ p∆V=nR∆T \\ ∆V>0=>thegasexpands \\ ∆T>0 \end{gathered}$$

Therefore,$∆U$ is positive because U increases when T increases.

$$\begin{gathered} \underset{~}{O}=W+∆U,W>0,∆U>0 \\ \underset{~}{O}>0 \end{gathered}$$

Hence heat flows into the gas.