Question:
One mole of an ideal monoatomic gas undergoes the process \(A \rightarrow B \rightarrow C \rightarrow D \rightarrow A\) as shown in the graph. The work done during the process is:
One mole of an ideal monoatomic gas undergoes the process \(A \rightarrow B \rightarrow C \rightarrow D \rightarrow A\) as shown in the graph. The work done during the process is:
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For a cyclic process on a \(P-V\) diagram, total work is the algebraic sum of work done in each process. Expansion gives positive work, while compression gives negative work.
Updated On: Jun 26, 2026
- \(-52.5\times 10^5\,\text{J}\)
- \(-11.5\times 10^5\,\text{J}\)
- \(-64\times 10^5\,\text{J}\)
- \(-36\times 10^5\,\text{J}\)
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The Correct Option is A
Solution and Explanation
Step 1: Identify the value of \(\gamma\).
For a monoatomic ideal gas, \[ \gamma=\frac{5}{3} \] Hence, \[ \gamma-1=\frac{5}{3}-1=\frac{2}{3} \]
Step 2: Find the coordinates of the points.
From the graph, \[ B=(1,1\times 10^5) \] and \[ C=(8,1\times 10^5) \] Also, \[ P_A=P_D=32\times 10^5 \] For adiabatic process, \[ PV^\gamma=\text{constant} \] For \(A \rightarrow B\), \[ P_AV_A^\gamma=P_BV_B^\gamma \] \[ 32\times 10^5 \times V_A^{5/3}=1\times 10^5 \times 1^{5/3} \] \[ 32V_A^{5/3}=1 \] \[ V_A^{5/3}=\frac{1}{32} \] Since, \[ 32=2^5 \] we get \[ V_A=\frac{1}{8} \] For \(C \rightarrow D\), \[ P_CV_C^\gamma=P_DV_D^\gamma \] \[ 1\times 10^5 \times 8^{5/3}=32\times 10^5 \times V_D^{5/3} \] Since, \[ 8^{5/3}=32 \] therefore, \[ 32=32V_D^{5/3} \] \[ V_D=1 \] So, \[ A=\left(\frac{1}{8},32\times 10^5\right),\quad B=(1,1\times 10^5) \] \[ C=(8,1\times 10^5),\quad D=(1,32\times 10^5) \]
Step 3: Calculate work done in adiabatic process \(A \rightarrow B\).
For an adiabatic process, \[ W=\frac{P_iV_i-P_fV_f}{\gamma-1} \] So, \[ W_{AB}=\frac{P_AV_A-P_BV_B}{\gamma-1} \] \[ W_{AB}=\frac{(32\times 10^5)(\frac{1}{8})-(1\times 10^5)(1)}{\frac{2}{3}} \] \[ W_{AB}=\frac{4\times 10^5-1\times 10^5}{\frac{2}{3}} \] \[ W_{AB}=\frac{3\times 10^5}{\frac{2}{3}} \] \[ W_{AB}=4.5\times 10^5\,\text{J} \]
Step 4: Calculate work done in isobaric process \(B \rightarrow C\).
For an isobaric process, \[ W=P\Delta V \] \[ W_{BC}=1\times 10^5(8-1) \] \[ W_{BC}=7\times 10^5\,\text{J} \]
Step 5: Calculate work done in adiabatic process \(C \rightarrow D\).
\[ W_{CD}=\frac{P_CV_C-P_DV_D}{\gamma-1} \] \[ W_{CD}=\frac{(1\times 10^5)(8)-(32\times 10^5)(1)}{\frac{2}{3}} \] \[ W_{CD}=\frac{8\times 10^5-32\times 10^5}{\frac{2}{3}} \] \[ W_{CD}=\frac{-24\times 10^5}{\frac{2}{3}} \] \[ W_{CD}=-36\times 10^5\,\text{J} \]
Step 6: Calculate work done in isobaric process \(D \rightarrow A\).
\[ W_{DA}=P(V_A-V_D) \] \[ W_{DA}=32\times 10^5\left(\frac{1}{8}-1\right) \] \[ W_{DA}=32\times 10^5\left(-\frac{7}{8}\right) \] \[ W_{DA}=-28\times 10^5\,\text{J} \]
Step 7: Find total work done.
\[ W_{\text{total}}=W_{AB}+W_{BC}+W_{CD}+W_{DA} \] \[ W_{\text{total}}=4.5\times 10^5+7\times 10^5-36\times 10^5-28\times 10^5 \] \[ W_{\text{total}}=-52.5\times 10^5\,\text{J} \]
Step 8: Final conclusion.
Therefore, \[ \boxed{-52.5\times 10^5\,\text{J}} \]
For a monoatomic ideal gas, \[ \gamma=\frac{5}{3} \] Hence, \[ \gamma-1=\frac{5}{3}-1=\frac{2}{3} \]
Step 2: Find the coordinates of the points.
From the graph, \[ B=(1,1\times 10^5) \] and \[ C=(8,1\times 10^5) \] Also, \[ P_A=P_D=32\times 10^5 \] For adiabatic process, \[ PV^\gamma=\text{constant} \] For \(A \rightarrow B\), \[ P_AV_A^\gamma=P_BV_B^\gamma \] \[ 32\times 10^5 \times V_A^{5/3}=1\times 10^5 \times 1^{5/3} \] \[ 32V_A^{5/3}=1 \] \[ V_A^{5/3}=\frac{1}{32} \] Since, \[ 32=2^5 \] we get \[ V_A=\frac{1}{8} \] For \(C \rightarrow D\), \[ P_CV_C^\gamma=P_DV_D^\gamma \] \[ 1\times 10^5 \times 8^{5/3}=32\times 10^5 \times V_D^{5/3} \] Since, \[ 8^{5/3}=32 \] therefore, \[ 32=32V_D^{5/3} \] \[ V_D=1 \] So, \[ A=\left(\frac{1}{8},32\times 10^5\right),\quad B=(1,1\times 10^5) \] \[ C=(8,1\times 10^5),\quad D=(1,32\times 10^5) \]
Step 3: Calculate work done in adiabatic process \(A \rightarrow B\).
For an adiabatic process, \[ W=\frac{P_iV_i-P_fV_f}{\gamma-1} \] So, \[ W_{AB}=\frac{P_AV_A-P_BV_B}{\gamma-1} \] \[ W_{AB}=\frac{(32\times 10^5)(\frac{1}{8})-(1\times 10^5)(1)}{\frac{2}{3}} \] \[ W_{AB}=\frac{4\times 10^5-1\times 10^5}{\frac{2}{3}} \] \[ W_{AB}=\frac{3\times 10^5}{\frac{2}{3}} \] \[ W_{AB}=4.5\times 10^5\,\text{J} \]
Step 4: Calculate work done in isobaric process \(B \rightarrow C\).
For an isobaric process, \[ W=P\Delta V \] \[ W_{BC}=1\times 10^5(8-1) \] \[ W_{BC}=7\times 10^5\,\text{J} \]
Step 5: Calculate work done in adiabatic process \(C \rightarrow D\).
\[ W_{CD}=\frac{P_CV_C-P_DV_D}{\gamma-1} \] \[ W_{CD}=\frac{(1\times 10^5)(8)-(32\times 10^5)(1)}{\frac{2}{3}} \] \[ W_{CD}=\frac{8\times 10^5-32\times 10^5}{\frac{2}{3}} \] \[ W_{CD}=\frac{-24\times 10^5}{\frac{2}{3}} \] \[ W_{CD}=-36\times 10^5\,\text{J} \]
Step 6: Calculate work done in isobaric process \(D \rightarrow A\).
\[ W_{DA}=P(V_A-V_D) \] \[ W_{DA}=32\times 10^5\left(\frac{1}{8}-1\right) \] \[ W_{DA}=32\times 10^5\left(-\frac{7}{8}\right) \] \[ W_{DA}=-28\times 10^5\,\text{J} \]
Step 7: Find total work done.
\[ W_{\text{total}}=W_{AB}+W_{BC}+W_{CD}+W_{DA} \] \[ W_{\text{total}}=4.5\times 10^5+7\times 10^5-36\times 10^5-28\times 10^5 \] \[ W_{\text{total}}=-52.5\times 10^5\,\text{J} \]
Step 8: Final conclusion.
Therefore, \[ \boxed{-52.5\times 10^5\,\text{J}} \]
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