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FIGURE shows two processes that take a gas from state i to
Chapter 17, Problem 57P(choose chapter or problem)
Figure P17.57 shows two processes that take a gas from state \(i\) to state \(f\). Show that \(Q_{A}-Q_{B}=p_{i} V_{i}\).
Equation Transcription:
Text Transcription:
i
f
Q_A-Q_B=p_iV_i
Questions & Answers
QUESTION:
Figure P17.57 shows two processes that take a gas from state \(i\) to state \(f\). Show that \(Q_{A}-Q_{B}=p_{i} V_{i}\).
Equation Transcription:
Text Transcription:
i
f
Q_A-Q_B=p_iV_i
ANSWER:Step 1 of 3
Our aim is to show the difference in the heat for the processes \(A\) and \(B\) is
\(Q_{A}-Q_{B}=P_{i} V_{i}\)
The initial pressure \(=P_{i}\)
The final pressure \(=2 \mathrm{P}_{\mathrm{i}}\)
The initial volume \(=V_{i}\)
The final volume \(=2 \mathrm{~V}_{\mathrm{i}}\)
The processes start from the point \(i\) and end at the point \(f\) finally. The changes in the energy of the two processes must be the same here. This is expressed as
\(\left(\Delta E_{y}\right)_{A}=\left(\Delta E_{y}\right)_{B}---(1)\)
The heat trans fer in the path \(\mathrm{A}\) is expressed as
\(Q_{A}=\left(\Delta E_{y}\right)_{A}-W_{A}---(2)\)
Where \(W_{A}\) is the work done during the path \(A\) It is nothing but the area under the curve. The system undergoes isochoric (constant volume) and isobaric process (linear expansion of volume) here. The work done is expressed as
\(W_{A}=0+\left(-2 P_{i}\right)\left(2 V_{i}-V_{i}\right)\)
\(W_{A}=-2 P_{i} V_{i}---(3)\)
Inserting the equation (3) in (2)
\(Q_{A}=\left(\Delta E_{y}\right)_{A}+2 P_{i} V_{i}---(4)\)