371276
One mole of an ideal monoatomic gas is taken round the cyclic process \({M N O M}\). The work done by the gas is
1 \({4.5 P_{0} V_{0}}\)
2 \({4 P_{0} V_{0}}\)
3 \({9 P_{0} V_{0}}\)
4 \({2 P_{0} V_{0}}\)
Explanation:
Work done for cyclic process is given by area of the loop. Work done \({W=\dfrac{1}{2} \times}\) Base \({\times}\) height \({=\dfrac{1}{2}\left(3 V_{0}-V_{0}\right)\left(3 P_{0}-P_{0}\right)}\) \({=\dfrac{1}{2}\left(2 V_{0}\right)\left(2 P_{0}\right)}\) \({W=2 P_{0} V_{0}}\).
KCET - 2024
PHXI12:THERMODYNAMICS
371277
The efficiency of an ideal gas with adiabatic exponent \(\gamma\) for the shown cyclic process would be
1 \(\dfrac{(2 \ln 2-1)}{\gamma /(\gamma-1)}\)
2 \(\dfrac{(1-1 \ln 2)}{\gamma /(\gamma-1)}\)
3 \(\dfrac{(2 \ln 2+1)}{\gamma /(\gamma-1)}\)
4 \(\dfrac{(2 \ln 2-1)}{\gamma /(\gamma+1)}\)
Explanation:
As, we know \(W=p \Delta V=n R \Delta T\) \(W_{B C}=-n R\left(2 T_{0}-T_{0}\right)=-n R T_{0}\) and \(W_{C A}=+2 n R T_{0} \ln 2\) Work done \(=W_{C A}+W_{B C}\) \( = 2\,nR{T_0}\ln 2 - n\,R{T_0}\) \( = n\,R{T_0}(2\ln 2 - 1)\) Also, input heat \(\Delta Q_{B C}=n C_{p} \Delta T=\dfrac{n R \gamma T_{0}}{\gamma-1}\) Efficiency \(=\dfrac{\text { Work }}{\text { Input heat }}=\dfrac{(2 \ln 2-1)}{\gamma /(\gamma-1)}\)
PHXI12:THERMODYNAMICS
371278
A thermodynamic system undergoes a cyclic process \(ABC\) as shown in the diagram. The work done by the system per cycle is
1 \( - 750\;J\)
2 \(750\;J\)
3 \(1250\;J\)
4 \( - 1250\;J\)
Explanation:
Work done \(=\) Area under \(P-V\) curve \( = \frac{1}{2} \times 300 \times 5 = 750\;J\) Since the work done by the system is in anticlockwise direction, work done will be negative i.e., \( - 750\;J\).
KCET - 2019
PHXI12:THERMODYNAMICS
371279
An engine operates by taking \(n\) moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is: (Take \(C_{V}=1.5 R\), where \(\mathrm{R}\) is gas constant)
1 0.24
2 0.15
3 0.32
4 0.08
Explanation:
The net work done in the cycle is equal to the area enclosed in the graph \(\Rightarrow W_{\text {cycle }}=P_{0} V_{0}\) \(Path\,AB\) \(\begin{aligned}\Delta Q_{A B} & =\Delta U+W \\& =\dfrac{P_{2} V_{2}-P_{1} V_{1}}{\gamma^{2}-1}+0=\dfrac{P_{0} V_{0}}{\dfrac{5}{3}-1}=\dfrac{3 P_{0} V_{0}}{2} \\\Delta Q_{B C} & =\Delta U+W \\& =\dfrac{4 P_{0} V_{0}-2 P_{0} V_{0}}{\dfrac{5}{3}-1}+2 P_{0} V_{0}=5 P_{0} V_{0}\end{aligned}\) In the paths \(CD\& DA,{\mkern 1mu} \,\Delta Q\) is (-) ve \(\eta = \frac{{{W_{cycle{\rm{ }}}}}}{{{Q_{absorbed{\rm{ }}}}}} = \frac{{{P_0}{V_0}}}{{\frac{{3{P_0}{V_0}}}{2} + 5{P_0}{V_0}}} = \frac{2}{{13}} = 0.15\)
371276
One mole of an ideal monoatomic gas is taken round the cyclic process \({M N O M}\). The work done by the gas is
1 \({4.5 P_{0} V_{0}}\)
2 \({4 P_{0} V_{0}}\)
3 \({9 P_{0} V_{0}}\)
4 \({2 P_{0} V_{0}}\)
Explanation:
Work done for cyclic process is given by area of the loop. Work done \({W=\dfrac{1}{2} \times}\) Base \({\times}\) height \({=\dfrac{1}{2}\left(3 V_{0}-V_{0}\right)\left(3 P_{0}-P_{0}\right)}\) \({=\dfrac{1}{2}\left(2 V_{0}\right)\left(2 P_{0}\right)}\) \({W=2 P_{0} V_{0}}\).
KCET - 2024
PHXI12:THERMODYNAMICS
371277
The efficiency of an ideal gas with adiabatic exponent \(\gamma\) for the shown cyclic process would be
1 \(\dfrac{(2 \ln 2-1)}{\gamma /(\gamma-1)}\)
2 \(\dfrac{(1-1 \ln 2)}{\gamma /(\gamma-1)}\)
3 \(\dfrac{(2 \ln 2+1)}{\gamma /(\gamma-1)}\)
4 \(\dfrac{(2 \ln 2-1)}{\gamma /(\gamma+1)}\)
Explanation:
As, we know \(W=p \Delta V=n R \Delta T\) \(W_{B C}=-n R\left(2 T_{0}-T_{0}\right)=-n R T_{0}\) and \(W_{C A}=+2 n R T_{0} \ln 2\) Work done \(=W_{C A}+W_{B C}\) \( = 2\,nR{T_0}\ln 2 - n\,R{T_0}\) \( = n\,R{T_0}(2\ln 2 - 1)\) Also, input heat \(\Delta Q_{B C}=n C_{p} \Delta T=\dfrac{n R \gamma T_{0}}{\gamma-1}\) Efficiency \(=\dfrac{\text { Work }}{\text { Input heat }}=\dfrac{(2 \ln 2-1)}{\gamma /(\gamma-1)}\)
PHXI12:THERMODYNAMICS
371278
A thermodynamic system undergoes a cyclic process \(ABC\) as shown in the diagram. The work done by the system per cycle is
1 \( - 750\;J\)
2 \(750\;J\)
3 \(1250\;J\)
4 \( - 1250\;J\)
Explanation:
Work done \(=\) Area under \(P-V\) curve \( = \frac{1}{2} \times 300 \times 5 = 750\;J\) Since the work done by the system is in anticlockwise direction, work done will be negative i.e., \( - 750\;J\).
KCET - 2019
PHXI12:THERMODYNAMICS
371279
An engine operates by taking \(n\) moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is: (Take \(C_{V}=1.5 R\), where \(\mathrm{R}\) is gas constant)
1 0.24
2 0.15
3 0.32
4 0.08
Explanation:
The net work done in the cycle is equal to the area enclosed in the graph \(\Rightarrow W_{\text {cycle }}=P_{0} V_{0}\) \(Path\,AB\) \(\begin{aligned}\Delta Q_{A B} & =\Delta U+W \\& =\dfrac{P_{2} V_{2}-P_{1} V_{1}}{\gamma^{2}-1}+0=\dfrac{P_{0} V_{0}}{\dfrac{5}{3}-1}=\dfrac{3 P_{0} V_{0}}{2} \\\Delta Q_{B C} & =\Delta U+W \\& =\dfrac{4 P_{0} V_{0}-2 P_{0} V_{0}}{\dfrac{5}{3}-1}+2 P_{0} V_{0}=5 P_{0} V_{0}\end{aligned}\) In the paths \(CD\& DA,{\mkern 1mu} \,\Delta Q\) is (-) ve \(\eta = \frac{{{W_{cycle{\rm{ }}}}}}{{{Q_{absorbed{\rm{ }}}}}} = \frac{{{P_0}{V_0}}}{{\frac{{3{P_0}{V_0}}}{2} + 5{P_0}{V_0}}} = \frac{2}{{13}} = 0.15\)
371276
One mole of an ideal monoatomic gas is taken round the cyclic process \({M N O M}\). The work done by the gas is
1 \({4.5 P_{0} V_{0}}\)
2 \({4 P_{0} V_{0}}\)
3 \({9 P_{0} V_{0}}\)
4 \({2 P_{0} V_{0}}\)
Explanation:
Work done for cyclic process is given by area of the loop. Work done \({W=\dfrac{1}{2} \times}\) Base \({\times}\) height \({=\dfrac{1}{2}\left(3 V_{0}-V_{0}\right)\left(3 P_{0}-P_{0}\right)}\) \({=\dfrac{1}{2}\left(2 V_{0}\right)\left(2 P_{0}\right)}\) \({W=2 P_{0} V_{0}}\).
KCET - 2024
PHXI12:THERMODYNAMICS
371277
The efficiency of an ideal gas with adiabatic exponent \(\gamma\) for the shown cyclic process would be
1 \(\dfrac{(2 \ln 2-1)}{\gamma /(\gamma-1)}\)
2 \(\dfrac{(1-1 \ln 2)}{\gamma /(\gamma-1)}\)
3 \(\dfrac{(2 \ln 2+1)}{\gamma /(\gamma-1)}\)
4 \(\dfrac{(2 \ln 2-1)}{\gamma /(\gamma+1)}\)
Explanation:
As, we know \(W=p \Delta V=n R \Delta T\) \(W_{B C}=-n R\left(2 T_{0}-T_{0}\right)=-n R T_{0}\) and \(W_{C A}=+2 n R T_{0} \ln 2\) Work done \(=W_{C A}+W_{B C}\) \( = 2\,nR{T_0}\ln 2 - n\,R{T_0}\) \( = n\,R{T_0}(2\ln 2 - 1)\) Also, input heat \(\Delta Q_{B C}=n C_{p} \Delta T=\dfrac{n R \gamma T_{0}}{\gamma-1}\) Efficiency \(=\dfrac{\text { Work }}{\text { Input heat }}=\dfrac{(2 \ln 2-1)}{\gamma /(\gamma-1)}\)
PHXI12:THERMODYNAMICS
371278
A thermodynamic system undergoes a cyclic process \(ABC\) as shown in the diagram. The work done by the system per cycle is
1 \( - 750\;J\)
2 \(750\;J\)
3 \(1250\;J\)
4 \( - 1250\;J\)
Explanation:
Work done \(=\) Area under \(P-V\) curve \( = \frac{1}{2} \times 300 \times 5 = 750\;J\) Since the work done by the system is in anticlockwise direction, work done will be negative i.e., \( - 750\;J\).
KCET - 2019
PHXI12:THERMODYNAMICS
371279
An engine operates by taking \(n\) moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is: (Take \(C_{V}=1.5 R\), where \(\mathrm{R}\) is gas constant)
1 0.24
2 0.15
3 0.32
4 0.08
Explanation:
The net work done in the cycle is equal to the area enclosed in the graph \(\Rightarrow W_{\text {cycle }}=P_{0} V_{0}\) \(Path\,AB\) \(\begin{aligned}\Delta Q_{A B} & =\Delta U+W \\& =\dfrac{P_{2} V_{2}-P_{1} V_{1}}{\gamma^{2}-1}+0=\dfrac{P_{0} V_{0}}{\dfrac{5}{3}-1}=\dfrac{3 P_{0} V_{0}}{2} \\\Delta Q_{B C} & =\Delta U+W \\& =\dfrac{4 P_{0} V_{0}-2 P_{0} V_{0}}{\dfrac{5}{3}-1}+2 P_{0} V_{0}=5 P_{0} V_{0}\end{aligned}\) In the paths \(CD\& DA,{\mkern 1mu} \,\Delta Q\) is (-) ve \(\eta = \frac{{{W_{cycle{\rm{ }}}}}}{{{Q_{absorbed{\rm{ }}}}}} = \frac{{{P_0}{V_0}}}{{\frac{{3{P_0}{V_0}}}{2} + 5{P_0}{V_0}}} = \frac{2}{{13}} = 0.15\)
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PHXI12:THERMODYNAMICS
371276
One mole of an ideal monoatomic gas is taken round the cyclic process \({M N O M}\). The work done by the gas is
1 \({4.5 P_{0} V_{0}}\)
2 \({4 P_{0} V_{0}}\)
3 \({9 P_{0} V_{0}}\)
4 \({2 P_{0} V_{0}}\)
Explanation:
Work done for cyclic process is given by area of the loop. Work done \({W=\dfrac{1}{2} \times}\) Base \({\times}\) height \({=\dfrac{1}{2}\left(3 V_{0}-V_{0}\right)\left(3 P_{0}-P_{0}\right)}\) \({=\dfrac{1}{2}\left(2 V_{0}\right)\left(2 P_{0}\right)}\) \({W=2 P_{0} V_{0}}\).
KCET - 2024
PHXI12:THERMODYNAMICS
371277
The efficiency of an ideal gas with adiabatic exponent \(\gamma\) for the shown cyclic process would be
1 \(\dfrac{(2 \ln 2-1)}{\gamma /(\gamma-1)}\)
2 \(\dfrac{(1-1 \ln 2)}{\gamma /(\gamma-1)}\)
3 \(\dfrac{(2 \ln 2+1)}{\gamma /(\gamma-1)}\)
4 \(\dfrac{(2 \ln 2-1)}{\gamma /(\gamma+1)}\)
Explanation:
As, we know \(W=p \Delta V=n R \Delta T\) \(W_{B C}=-n R\left(2 T_{0}-T_{0}\right)=-n R T_{0}\) and \(W_{C A}=+2 n R T_{0} \ln 2\) Work done \(=W_{C A}+W_{B C}\) \( = 2\,nR{T_0}\ln 2 - n\,R{T_0}\) \( = n\,R{T_0}(2\ln 2 - 1)\) Also, input heat \(\Delta Q_{B C}=n C_{p} \Delta T=\dfrac{n R \gamma T_{0}}{\gamma-1}\) Efficiency \(=\dfrac{\text { Work }}{\text { Input heat }}=\dfrac{(2 \ln 2-1)}{\gamma /(\gamma-1)}\)
PHXI12:THERMODYNAMICS
371278
A thermodynamic system undergoes a cyclic process \(ABC\) as shown in the diagram. The work done by the system per cycle is
1 \( - 750\;J\)
2 \(750\;J\)
3 \(1250\;J\)
4 \( - 1250\;J\)
Explanation:
Work done \(=\) Area under \(P-V\) curve \( = \frac{1}{2} \times 300 \times 5 = 750\;J\) Since the work done by the system is in anticlockwise direction, work done will be negative i.e., \( - 750\;J\).
KCET - 2019
PHXI12:THERMODYNAMICS
371279
An engine operates by taking \(n\) moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is: (Take \(C_{V}=1.5 R\), where \(\mathrm{R}\) is gas constant)
1 0.24
2 0.15
3 0.32
4 0.08
Explanation:
The net work done in the cycle is equal to the area enclosed in the graph \(\Rightarrow W_{\text {cycle }}=P_{0} V_{0}\) \(Path\,AB\) \(\begin{aligned}\Delta Q_{A B} & =\Delta U+W \\& =\dfrac{P_{2} V_{2}-P_{1} V_{1}}{\gamma^{2}-1}+0=\dfrac{P_{0} V_{0}}{\dfrac{5}{3}-1}=\dfrac{3 P_{0} V_{0}}{2} \\\Delta Q_{B C} & =\Delta U+W \\& =\dfrac{4 P_{0} V_{0}-2 P_{0} V_{0}}{\dfrac{5}{3}-1}+2 P_{0} V_{0}=5 P_{0} V_{0}\end{aligned}\) In the paths \(CD\& DA,{\mkern 1mu} \,\Delta Q\) is (-) ve \(\eta = \frac{{{W_{cycle{\rm{ }}}}}}{{{Q_{absorbed{\rm{ }}}}}} = \frac{{{P_0}{V_0}}}{{\frac{{3{P_0}{V_0}}}{2} + 5{P_0}{V_0}}} = \frac{2}{{13}} = 0.15\)
