164177
\(\quad V=k\left(\frac{P}{T}\right)^{0.33}\) where \(k\) is constant, it is :
1 isothermal process
2 adiabatic process
3 isochoric process
4 isobaric process
Explanation:
\(\mathrm{V}=\mathrm{K} \cdot\left(\frac{\mathrm{P}}{\mathrm{T}}\right)^{0.33}, \mathrm{~K}=\) constant,\(\left(\frac{\mathrm{P}}{\mathrm{T}}\right)=\frac{\mathrm{nR}}{\mathrm{V}}\) so, \(\mathrm{V}=\mathrm{K} \cdot\left(\frac{\mathrm{nR}}{\mathrm{V}}\right)^{1 / 3} \Rightarrow \mathrm{V}^{2 / 3}\) Isochoric process
NCERT-XI-II-235
7 RBTS PAPER
164178
Six moles of an ideal gas performs a cycle shown in figure. If the temperature are \(T_A=600 \mathrm{~K}\), \(T_B=800 \mathrm{~K}, T_C=2200 K\) and \(T_D=1200 \mathrm{~K}\), the work done per cycle is:
1 \(20000 \mathrm{~J}\)
2 \(10000 \mathrm{~J}\)
3 \(30000 \mathrm{~J}\)
4 \(40000 \mathrm{~J}\)
Explanation:
Work done \(\mathrm{B}\) to \(\mathrm{C}=\mathrm{P}_2\left(\mathrm{~V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}\right)\) \( =\mu R\left(T_C-T_B\right) \) \( =6 \times R \times(2200-800) \) \( =6 R \times 1400 \)
Work done \(\mathrm{D}\) to \(\mathrm{A} \quad=\mathrm{P}_1\left(\mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}\right)\) \( =6 R(600-1200) \) \( =-6 R(600) \)
Work done from \(A\) to \(B\) and \(B\) to \(C\) is zero because of constant volume \(\therefore\) Total work done \( =6 \mathrm{R} \times 1400-6 \mathrm{R} \times 800 \) \( =6 \mathrm{R}(800) \) \( =50 \times 800=40000 \mathrm{~J} \)
NCERT-XI-II-235
7 RBTS PAPER
164179
Change in internal energy of a system :
1 depends only on initial and final states of the system
2 does not depend on path of the process
3 is zero during a cyclic process
4 all of the above statements are correct
Explanation:
Depends only on initial and final states of the system, does not depend on path of the process is zero during a cyclic process.
NCERT-XI-II-230
7 RBTS PAPER
164180
First law of thermodynamics is statement of conservation of :
1 Mass
2 Energy
3 Linear momentum
4 Angular momentum
Explanation:
It is based on energy conservation.
NCERT-XI-II-230
7 RBTS PAPER
164181
In the cyclic process \(A B C A\), as shown in the adjacent figure, the work done during processes \(A \rightarrow B, B \rightarrow C\) and \(C \rightarrow A\); are :
164177
\(\quad V=k\left(\frac{P}{T}\right)^{0.33}\) where \(k\) is constant, it is :
1 isothermal process
2 adiabatic process
3 isochoric process
4 isobaric process
Explanation:
\(\mathrm{V}=\mathrm{K} \cdot\left(\frac{\mathrm{P}}{\mathrm{T}}\right)^{0.33}, \mathrm{~K}=\) constant,\(\left(\frac{\mathrm{P}}{\mathrm{T}}\right)=\frac{\mathrm{nR}}{\mathrm{V}}\) so, \(\mathrm{V}=\mathrm{K} \cdot\left(\frac{\mathrm{nR}}{\mathrm{V}}\right)^{1 / 3} \Rightarrow \mathrm{V}^{2 / 3}\) Isochoric process
NCERT-XI-II-235
7 RBTS PAPER
164178
Six moles of an ideal gas performs a cycle shown in figure. If the temperature are \(T_A=600 \mathrm{~K}\), \(T_B=800 \mathrm{~K}, T_C=2200 K\) and \(T_D=1200 \mathrm{~K}\), the work done per cycle is:
1 \(20000 \mathrm{~J}\)
2 \(10000 \mathrm{~J}\)
3 \(30000 \mathrm{~J}\)
4 \(40000 \mathrm{~J}\)
Explanation:
Work done \(\mathrm{B}\) to \(\mathrm{C}=\mathrm{P}_2\left(\mathrm{~V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}\right)\) \( =\mu R\left(T_C-T_B\right) \) \( =6 \times R \times(2200-800) \) \( =6 R \times 1400 \)
Work done \(\mathrm{D}\) to \(\mathrm{A} \quad=\mathrm{P}_1\left(\mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}\right)\) \( =6 R(600-1200) \) \( =-6 R(600) \)
Work done from \(A\) to \(B\) and \(B\) to \(C\) is zero because of constant volume \(\therefore\) Total work done \( =6 \mathrm{R} \times 1400-6 \mathrm{R} \times 800 \) \( =6 \mathrm{R}(800) \) \( =50 \times 800=40000 \mathrm{~J} \)
NCERT-XI-II-235
7 RBTS PAPER
164179
Change in internal energy of a system :
1 depends only on initial and final states of the system
2 does not depend on path of the process
3 is zero during a cyclic process
4 all of the above statements are correct
Explanation:
Depends only on initial and final states of the system, does not depend on path of the process is zero during a cyclic process.
NCERT-XI-II-230
7 RBTS PAPER
164180
First law of thermodynamics is statement of conservation of :
1 Mass
2 Energy
3 Linear momentum
4 Angular momentum
Explanation:
It is based on energy conservation.
NCERT-XI-II-230
7 RBTS PAPER
164181
In the cyclic process \(A B C A\), as shown in the adjacent figure, the work done during processes \(A \rightarrow B, B \rightarrow C\) and \(C \rightarrow A\); are :
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7 RBTS PAPER
164177
\(\quad V=k\left(\frac{P}{T}\right)^{0.33}\) where \(k\) is constant, it is :
1 isothermal process
2 adiabatic process
3 isochoric process
4 isobaric process
Explanation:
\(\mathrm{V}=\mathrm{K} \cdot\left(\frac{\mathrm{P}}{\mathrm{T}}\right)^{0.33}, \mathrm{~K}=\) constant,\(\left(\frac{\mathrm{P}}{\mathrm{T}}\right)=\frac{\mathrm{nR}}{\mathrm{V}}\) so, \(\mathrm{V}=\mathrm{K} \cdot\left(\frac{\mathrm{nR}}{\mathrm{V}}\right)^{1 / 3} \Rightarrow \mathrm{V}^{2 / 3}\) Isochoric process
NCERT-XI-II-235
7 RBTS PAPER
164178
Six moles of an ideal gas performs a cycle shown in figure. If the temperature are \(T_A=600 \mathrm{~K}\), \(T_B=800 \mathrm{~K}, T_C=2200 K\) and \(T_D=1200 \mathrm{~K}\), the work done per cycle is:
1 \(20000 \mathrm{~J}\)
2 \(10000 \mathrm{~J}\)
3 \(30000 \mathrm{~J}\)
4 \(40000 \mathrm{~J}\)
Explanation:
Work done \(\mathrm{B}\) to \(\mathrm{C}=\mathrm{P}_2\left(\mathrm{~V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}\right)\) \( =\mu R\left(T_C-T_B\right) \) \( =6 \times R \times(2200-800) \) \( =6 R \times 1400 \)
Work done \(\mathrm{D}\) to \(\mathrm{A} \quad=\mathrm{P}_1\left(\mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}\right)\) \( =6 R(600-1200) \) \( =-6 R(600) \)
Work done from \(A\) to \(B\) and \(B\) to \(C\) is zero because of constant volume \(\therefore\) Total work done \( =6 \mathrm{R} \times 1400-6 \mathrm{R} \times 800 \) \( =6 \mathrm{R}(800) \) \( =50 \times 800=40000 \mathrm{~J} \)
NCERT-XI-II-235
7 RBTS PAPER
164179
Change in internal energy of a system :
1 depends only on initial and final states of the system
2 does not depend on path of the process
3 is zero during a cyclic process
4 all of the above statements are correct
Explanation:
Depends only on initial and final states of the system, does not depend on path of the process is zero during a cyclic process.
NCERT-XI-II-230
7 RBTS PAPER
164180
First law of thermodynamics is statement of conservation of :
1 Mass
2 Energy
3 Linear momentum
4 Angular momentum
Explanation:
It is based on energy conservation.
NCERT-XI-II-230
7 RBTS PAPER
164181
In the cyclic process \(A B C A\), as shown in the adjacent figure, the work done during processes \(A \rightarrow B, B \rightarrow C\) and \(C \rightarrow A\); are :
164177
\(\quad V=k\left(\frac{P}{T}\right)^{0.33}\) where \(k\) is constant, it is :
1 isothermal process
2 adiabatic process
3 isochoric process
4 isobaric process
Explanation:
\(\mathrm{V}=\mathrm{K} \cdot\left(\frac{\mathrm{P}}{\mathrm{T}}\right)^{0.33}, \mathrm{~K}=\) constant,\(\left(\frac{\mathrm{P}}{\mathrm{T}}\right)=\frac{\mathrm{nR}}{\mathrm{V}}\) so, \(\mathrm{V}=\mathrm{K} \cdot\left(\frac{\mathrm{nR}}{\mathrm{V}}\right)^{1 / 3} \Rightarrow \mathrm{V}^{2 / 3}\) Isochoric process
NCERT-XI-II-235
7 RBTS PAPER
164178
Six moles of an ideal gas performs a cycle shown in figure. If the temperature are \(T_A=600 \mathrm{~K}\), \(T_B=800 \mathrm{~K}, T_C=2200 K\) and \(T_D=1200 \mathrm{~K}\), the work done per cycle is:
1 \(20000 \mathrm{~J}\)
2 \(10000 \mathrm{~J}\)
3 \(30000 \mathrm{~J}\)
4 \(40000 \mathrm{~J}\)
Explanation:
Work done \(\mathrm{B}\) to \(\mathrm{C}=\mathrm{P}_2\left(\mathrm{~V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}\right)\) \( =\mu R\left(T_C-T_B\right) \) \( =6 \times R \times(2200-800) \) \( =6 R \times 1400 \)
Work done \(\mathrm{D}\) to \(\mathrm{A} \quad=\mathrm{P}_1\left(\mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}\right)\) \( =6 R(600-1200) \) \( =-6 R(600) \)
Work done from \(A\) to \(B\) and \(B\) to \(C\) is zero because of constant volume \(\therefore\) Total work done \( =6 \mathrm{R} \times 1400-6 \mathrm{R} \times 800 \) \( =6 \mathrm{R}(800) \) \( =50 \times 800=40000 \mathrm{~J} \)
NCERT-XI-II-235
7 RBTS PAPER
164179
Change in internal energy of a system :
1 depends only on initial and final states of the system
2 does not depend on path of the process
3 is zero during a cyclic process
4 all of the above statements are correct
Explanation:
Depends only on initial and final states of the system, does not depend on path of the process is zero during a cyclic process.
NCERT-XI-II-230
7 RBTS PAPER
164180
First law of thermodynamics is statement of conservation of :
1 Mass
2 Energy
3 Linear momentum
4 Angular momentum
Explanation:
It is based on energy conservation.
NCERT-XI-II-230
7 RBTS PAPER
164181
In the cyclic process \(A B C A\), as shown in the adjacent figure, the work done during processes \(A \rightarrow B, B \rightarrow C\) and \(C \rightarrow A\); are :
164177
\(\quad V=k\left(\frac{P}{T}\right)^{0.33}\) where \(k\) is constant, it is :
1 isothermal process
2 adiabatic process
3 isochoric process
4 isobaric process
Explanation:
\(\mathrm{V}=\mathrm{K} \cdot\left(\frac{\mathrm{P}}{\mathrm{T}}\right)^{0.33}, \mathrm{~K}=\) constant,\(\left(\frac{\mathrm{P}}{\mathrm{T}}\right)=\frac{\mathrm{nR}}{\mathrm{V}}\) so, \(\mathrm{V}=\mathrm{K} \cdot\left(\frac{\mathrm{nR}}{\mathrm{V}}\right)^{1 / 3} \Rightarrow \mathrm{V}^{2 / 3}\) Isochoric process
NCERT-XI-II-235
7 RBTS PAPER
164178
Six moles of an ideal gas performs a cycle shown in figure. If the temperature are \(T_A=600 \mathrm{~K}\), \(T_B=800 \mathrm{~K}, T_C=2200 K\) and \(T_D=1200 \mathrm{~K}\), the work done per cycle is:
1 \(20000 \mathrm{~J}\)
2 \(10000 \mathrm{~J}\)
3 \(30000 \mathrm{~J}\)
4 \(40000 \mathrm{~J}\)
Explanation:
Work done \(\mathrm{B}\) to \(\mathrm{C}=\mathrm{P}_2\left(\mathrm{~V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}\right)\) \( =\mu R\left(T_C-T_B\right) \) \( =6 \times R \times(2200-800) \) \( =6 R \times 1400 \)
Work done \(\mathrm{D}\) to \(\mathrm{A} \quad=\mathrm{P}_1\left(\mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}\right)\) \( =6 R(600-1200) \) \( =-6 R(600) \)
Work done from \(A\) to \(B\) and \(B\) to \(C\) is zero because of constant volume \(\therefore\) Total work done \( =6 \mathrm{R} \times 1400-6 \mathrm{R} \times 800 \) \( =6 \mathrm{R}(800) \) \( =50 \times 800=40000 \mathrm{~J} \)
NCERT-XI-II-235
7 RBTS PAPER
164179
Change in internal energy of a system :
1 depends only on initial and final states of the system
2 does not depend on path of the process
3 is zero during a cyclic process
4 all of the above statements are correct
Explanation:
Depends only on initial and final states of the system, does not depend on path of the process is zero during a cyclic process.
NCERT-XI-II-230
7 RBTS PAPER
164180
First law of thermodynamics is statement of conservation of :
1 Mass
2 Energy
3 Linear momentum
4 Angular momentum
Explanation:
It is based on energy conservation.
NCERT-XI-II-230
7 RBTS PAPER
164181
In the cyclic process \(A B C A\), as shown in the adjacent figure, the work done during processes \(A \rightarrow B, B \rightarrow C\) and \(C \rightarrow A\); are :
