Does the temperature of an ideal gas increase, decrease, or stay the same during (a) an isothermal expansion, (b) an expansion at constant pressure, (c) an adiabatic expansion, and (d) an increase in pressure at constant volume?

The gas is an ideal gas.

Since the temperature remains constant in an isothermal process, the temperature of an ideal gas during an isothermal expansion stays the same. Using the gas law, we can predict the temperature of an ideal gas during an expansion at constant pressure and during an increase in pressure at constant volume. Then, using the first law of thermodynamics, we can predict the temperature of an ideal gas during an adiabatic expansion.

Formulae:

Ideal gas equation, $pV=nRT$ …(i)

Change in internal energy due to first law of thermodynamics,$∆E_{int}=Q-W$…(ii)

Change in internal energy at constant volume, $∆E_{int}=n{C}_v∆T$…(iii)

For an ideal gas at isothermal expansion, the initial temperature and final temperature is the same, so$\Delta T=0.$

Hence, the temperature of an ideal gas during an isothermal expansion stays the same.

For constant pressure, there is change in volume. Thus, the equation (i) gives the temperature value as follows:

$\begin{array}{c}p\Delta V=nR\Delta T\\ \Delta T=\frac{p\Delta V}{nR}\end{array}$

Since, $\Delta V>0$thus, the value of change in temperature using the above equation is$\Delta T>0$

Hence, the change in temperature increases at constant pressure.

For adiabatic expansion, the total heat is constant, i.e.$Q=0$

Thus, using equation (ii), the change in internal energy is given as follows:

$∆E_{int}=-W$

For adiabatic expansion, gas is expanding, so that work done is positive and is increasing.

Thus,$∆E_{int}<0$

Now, using this value in equation (iii), we can get the change in temperature as follows:

$\Delta T<0$

Hence, the change in temperature decreases at constant adiabatic expansion.

For constant volume, there is change in pressure. Thus, equation (i) gives the temperature value as follows:

$\begin{array}{c}\Delta pV=nR\Delta T\\ \Delta T=\frac{\Delta pV}{nR}\end{array}$

Since, $\Delta p>0$the change in internal energy is found using the above equation as$\Delta T>0$

Therefore, the change in temperature increases with an increase in the pressure at constant volume.