148235
The thermodynamic process in which no work is done on or by the gas is
1 isothermal process
2 adiabatic process
3 cyclic process
4 isobaric process
5 isochoric process
Explanation:
E For isochoric process, $\mathrm{V}=$ constant. $\mathrm{W}=\mathrm{PdV}$ $\mathrm{W}=0$
Kerala CEE - 2011
Thermodynamics
148236
During the adiabatic expansion of $2 \mathrm{~mol}$ of an ideal gas, the increase internal energy was found to be equal to $(-200 \mathrm{~J})$. The work done by the gas during the process will be equal to
148237
A system undergoes a reversible adiabatic process. The entropy of the system
1 remains constant
2 may increase or may decrease
3 increases
4 decreases
Explanation:
A For Adiabatic process, $\Delta \mathrm{Q}=0$ We know, $\Delta \mathrm{S} =\frac{\Delta \mathrm{Q}}{\mathrm{T}} \quad(\because \Delta \mathrm{Q}=0)$ $\Delta \mathrm{S} =0,$ $\mathrm{~S} =\mathrm{const}$ Hence, entropy of the system remains constant .
UPSEE - 2016
Thermodynamics
148238
If one mole of a monoatomic gas $\left(\gamma_{1}=\frac{5}{3}\right)$ is mixed with one mole of a diatomic gas $\left(\gamma_{2}=\frac{7}{5}\right)$, the value of $\gamma$ for the mixture is
1 3.07
2 1.5
3 1.40
4 1.53
Explanation:
B For monoatomic gas, $\gamma_{1}=\frac{5}{3}, \quad C_{V}=\frac{3}{2} R \text { and } C_{P}=\frac{5}{2} R$ For diatomic gas, $\gamma_{2}=\frac{7}{5}, \quad \mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R} \text { and } \mathrm{C}_{\mathrm{P}}=\frac{7}{2} \mathrm{R}$ $\left(\mathrm{C}_{\mathrm{V}}\right)_{\mathrm{mix}}=\frac{\mathrm{n}_{1} \mathrm{C}_{\mathrm{V}_{1}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{V}_{2}}}{\mathrm{n}_{1}+\mathrm{n}_{2}}=\frac{1 \times\left(\frac{3}{2} R\right)+1\left(\frac{5}{2} R\right)}{1+1}$ $=\frac{8 \mathrm{R}}{2 \times 2}=2 \mathrm{R}$ \(\left(\mathrm{C}_{\mathrm{V}}\right)_{\text {mix }}=2 \mathrm{R}\) \(\text { Similarly }\left(\mathrm{C}_{\mathrm{p}}\right)=3 \mathrm{R}\) \(\gamma_{\text {mixture }}=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{3 \mathrm{R}}{2 \mathrm{R}}=1.5\) \(\gamma_{\text {mixture }}=1.5\)
148235
The thermodynamic process in which no work is done on or by the gas is
1 isothermal process
2 adiabatic process
3 cyclic process
4 isobaric process
5 isochoric process
Explanation:
E For isochoric process, $\mathrm{V}=$ constant. $\mathrm{W}=\mathrm{PdV}$ $\mathrm{W}=0$
Kerala CEE - 2011
Thermodynamics
148236
During the adiabatic expansion of $2 \mathrm{~mol}$ of an ideal gas, the increase internal energy was found to be equal to $(-200 \mathrm{~J})$. The work done by the gas during the process will be equal to
148237
A system undergoes a reversible adiabatic process. The entropy of the system
1 remains constant
2 may increase or may decrease
3 increases
4 decreases
Explanation:
A For Adiabatic process, $\Delta \mathrm{Q}=0$ We know, $\Delta \mathrm{S} =\frac{\Delta \mathrm{Q}}{\mathrm{T}} \quad(\because \Delta \mathrm{Q}=0)$ $\Delta \mathrm{S} =0,$ $\mathrm{~S} =\mathrm{const}$ Hence, entropy of the system remains constant .
UPSEE - 2016
Thermodynamics
148238
If one mole of a monoatomic gas $\left(\gamma_{1}=\frac{5}{3}\right)$ is mixed with one mole of a diatomic gas $\left(\gamma_{2}=\frac{7}{5}\right)$, the value of $\gamma$ for the mixture is
1 3.07
2 1.5
3 1.40
4 1.53
Explanation:
B For monoatomic gas, $\gamma_{1}=\frac{5}{3}, \quad C_{V}=\frac{3}{2} R \text { and } C_{P}=\frac{5}{2} R$ For diatomic gas, $\gamma_{2}=\frac{7}{5}, \quad \mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R} \text { and } \mathrm{C}_{\mathrm{P}}=\frac{7}{2} \mathrm{R}$ $\left(\mathrm{C}_{\mathrm{V}}\right)_{\mathrm{mix}}=\frac{\mathrm{n}_{1} \mathrm{C}_{\mathrm{V}_{1}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{V}_{2}}}{\mathrm{n}_{1}+\mathrm{n}_{2}}=\frac{1 \times\left(\frac{3}{2} R\right)+1\left(\frac{5}{2} R\right)}{1+1}$ $=\frac{8 \mathrm{R}}{2 \times 2}=2 \mathrm{R}$ \(\left(\mathrm{C}_{\mathrm{V}}\right)_{\text {mix }}=2 \mathrm{R}\) \(\text { Similarly }\left(\mathrm{C}_{\mathrm{p}}\right)=3 \mathrm{R}\) \(\gamma_{\text {mixture }}=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{3 \mathrm{R}}{2 \mathrm{R}}=1.5\) \(\gamma_{\text {mixture }}=1.5\)
148235
The thermodynamic process in which no work is done on or by the gas is
1 isothermal process
2 adiabatic process
3 cyclic process
4 isobaric process
5 isochoric process
Explanation:
E For isochoric process, $\mathrm{V}=$ constant. $\mathrm{W}=\mathrm{PdV}$ $\mathrm{W}=0$
Kerala CEE - 2011
Thermodynamics
148236
During the adiabatic expansion of $2 \mathrm{~mol}$ of an ideal gas, the increase internal energy was found to be equal to $(-200 \mathrm{~J})$. The work done by the gas during the process will be equal to
148237
A system undergoes a reversible adiabatic process. The entropy of the system
1 remains constant
2 may increase or may decrease
3 increases
4 decreases
Explanation:
A For Adiabatic process, $\Delta \mathrm{Q}=0$ We know, $\Delta \mathrm{S} =\frac{\Delta \mathrm{Q}}{\mathrm{T}} \quad(\because \Delta \mathrm{Q}=0)$ $\Delta \mathrm{S} =0,$ $\mathrm{~S} =\mathrm{const}$ Hence, entropy of the system remains constant .
UPSEE - 2016
Thermodynamics
148238
If one mole of a monoatomic gas $\left(\gamma_{1}=\frac{5}{3}\right)$ is mixed with one mole of a diatomic gas $\left(\gamma_{2}=\frac{7}{5}\right)$, the value of $\gamma$ for the mixture is
1 3.07
2 1.5
3 1.40
4 1.53
Explanation:
B For monoatomic gas, $\gamma_{1}=\frac{5}{3}, \quad C_{V}=\frac{3}{2} R \text { and } C_{P}=\frac{5}{2} R$ For diatomic gas, $\gamma_{2}=\frac{7}{5}, \quad \mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R} \text { and } \mathrm{C}_{\mathrm{P}}=\frac{7}{2} \mathrm{R}$ $\left(\mathrm{C}_{\mathrm{V}}\right)_{\mathrm{mix}}=\frac{\mathrm{n}_{1} \mathrm{C}_{\mathrm{V}_{1}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{V}_{2}}}{\mathrm{n}_{1}+\mathrm{n}_{2}}=\frac{1 \times\left(\frac{3}{2} R\right)+1\left(\frac{5}{2} R\right)}{1+1}$ $=\frac{8 \mathrm{R}}{2 \times 2}=2 \mathrm{R}$ \(\left(\mathrm{C}_{\mathrm{V}}\right)_{\text {mix }}=2 \mathrm{R}\) \(\text { Similarly }\left(\mathrm{C}_{\mathrm{p}}\right)=3 \mathrm{R}\) \(\gamma_{\text {mixture }}=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{3 \mathrm{R}}{2 \mathrm{R}}=1.5\) \(\gamma_{\text {mixture }}=1.5\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
148235
The thermodynamic process in which no work is done on or by the gas is
1 isothermal process
2 adiabatic process
3 cyclic process
4 isobaric process
5 isochoric process
Explanation:
E For isochoric process, $\mathrm{V}=$ constant. $\mathrm{W}=\mathrm{PdV}$ $\mathrm{W}=0$
Kerala CEE - 2011
Thermodynamics
148236
During the adiabatic expansion of $2 \mathrm{~mol}$ of an ideal gas, the increase internal energy was found to be equal to $(-200 \mathrm{~J})$. The work done by the gas during the process will be equal to
148237
A system undergoes a reversible adiabatic process. The entropy of the system
1 remains constant
2 may increase or may decrease
3 increases
4 decreases
Explanation:
A For Adiabatic process, $\Delta \mathrm{Q}=0$ We know, $\Delta \mathrm{S} =\frac{\Delta \mathrm{Q}}{\mathrm{T}} \quad(\because \Delta \mathrm{Q}=0)$ $\Delta \mathrm{S} =0,$ $\mathrm{~S} =\mathrm{const}$ Hence, entropy of the system remains constant .
UPSEE - 2016
Thermodynamics
148238
If one mole of a monoatomic gas $\left(\gamma_{1}=\frac{5}{3}\right)$ is mixed with one mole of a diatomic gas $\left(\gamma_{2}=\frac{7}{5}\right)$, the value of $\gamma$ for the mixture is
1 3.07
2 1.5
3 1.40
4 1.53
Explanation:
B For monoatomic gas, $\gamma_{1}=\frac{5}{3}, \quad C_{V}=\frac{3}{2} R \text { and } C_{P}=\frac{5}{2} R$ For diatomic gas, $\gamma_{2}=\frac{7}{5}, \quad \mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R} \text { and } \mathrm{C}_{\mathrm{P}}=\frac{7}{2} \mathrm{R}$ $\left(\mathrm{C}_{\mathrm{V}}\right)_{\mathrm{mix}}=\frac{\mathrm{n}_{1} \mathrm{C}_{\mathrm{V}_{1}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{V}_{2}}}{\mathrm{n}_{1}+\mathrm{n}_{2}}=\frac{1 \times\left(\frac{3}{2} R\right)+1\left(\frac{5}{2} R\right)}{1+1}$ $=\frac{8 \mathrm{R}}{2 \times 2}=2 \mathrm{R}$ \(\left(\mathrm{C}_{\mathrm{V}}\right)_{\text {mix }}=2 \mathrm{R}\) \(\text { Similarly }\left(\mathrm{C}_{\mathrm{p}}\right)=3 \mathrm{R}\) \(\gamma_{\text {mixture }}=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{3 \mathrm{R}}{2 \mathrm{R}}=1.5\) \(\gamma_{\text {mixture }}=1.5\)
