Solved Problem on Thermodynamics

A gas undergoes two sucessive processes, first, an isometric process, where it receives 200 joules of heat, then an isobaric process receiving 150 joules of heat, as show in the figure. Calculate in each process the work done in the process and the change in internal energy of the gas.

Problem data:

Heat received in isometric pocess: Q_AB = 200 J;
Heat received in isobaric process: Q_BC = 150 J.

Solution

Isometric, isochoric, or isovolumetric process from A to B

In this process, there is no change in volume, ΔV = 0, therefore the work done will be zero

\[ \begin{gather} \bbox[#99CCFF,10px] {W_{A}^{B}=P\Delta V} \tag{I} \end{gather} \]

\[ W_{A}^{B}=P \times 0 \]

\[ \bbox[#FFCCCC,10px] {W_{A}^{B}=0} \]

The change of internal energy in this process will be

\[ \begin{gather} \bbox[#99CCFF,10px] {\Delta U_{AB}=Q_{AB}-W_{A}^{B}} \tag{II} \end{gather} \]

\[ \begin{gather} \Delta U_{AB}=Q_{AB}-0\\ \Delta U_{AB}=Q_{AB} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {\Delta U_{AB}=200\;\text{J}} \]

Isobaric process from B to C

Applying the expression (I) to the process between B and C, we have

\[ \begin{gather} W_{B}^{C}=P (V_{C}-V_{B})\\ W_{B}^{C}=5\times (30-10)\\ W_{B}^{C}=5\times 20 \end{gather} \]

\[ \bbox[#FFCCCC,10px] {W_{B}^{C}=100\;\text{J}} \]

Applying the expression (II), we have

\[ \begin{gather} \Delta U_{BC}=Q_{BC}-W_{B}^{C}\\ \Delta U_{BC}=150-100 \end{gather} \]

\[ \bbox[#FFCCCC,10px] {\Delta U_{BC}=50\;\text{J}} \]

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