A gas under goes the process shown in FIGURE Q18.10. By what factor does the temperature change?

The pV diagram of the gas undergoes isobaric processes.

Visualize: The isobaric process of gas is shown in the figure. It shows that the pressure of gas remains the same and the volume goes three times of its initial.

Here;

$$\begin{gathered} P_i=InitialPressure \\ {V}_i=InitialVolume \\ {T}_i=InitialTemparture \end{gathered}$$

respectably the initial pressure, volume and temperature of the gas.

Fromideal Gas Equation,

If, $T_i=\frac{P_i{V}_i}{nR}$are the final parameters

Here $P_f,V_f,\&T_f$are the final parameters

$T_f=\frac{P_f{V}_f}{nR}$

According to the graph, pressure is seen to be remaining constant and volume is increasing three times, so;

$P_i=P_f$and $V_f=3V_i$

Final Temperature of the equation is thus seen to be;

$\begin{array}{rcl}{T}_f& =& \frac{P_i\left(3V_i\right)}{nR}\\ & =& 3\left(\frac{P_i{V}_i}{nR}\right)\\ T_f& =& 3T_i\end{array}$

The final temperature of the gas will be $3Times$of the initial temperature of the gas.