139319
Specific heat of a gas undergoing adiabatic change is
1 Zero
2 Infinite
3 Positive
4 Negative
Explanation:
A Specific heat of a gas undergoing adiabatic change is zero because during adiabatic process no heat is supplied to system.
AP EAMCET-23.08.2021
Kinetic Theory of Gases
139323
For an ideal gas, if the ratio of Molar specific heats $\gamma=1.4$, then the specific heat at constant pressure $C_{P}$, specific heat at constant volume $C_{V}$ and corresponding molecule are respectively
C Ratio of molar specific heat $\gamma=1.4$ $\frac{C_{P}}{C_{V}}=\gamma=1.4$ $\frac{C_{P}}{C_{V}}=\frac{14}{10}=\frac{7}{5} \quad\left(\because \frac{C_{P}}{C_{V}}=\frac{7}{5}\right.$ so, diatomic rigid $)$ $C_{V}=\frac{f}{2} R$ So, $C_{V}=\frac{5}{2} R \quad\left(\because \gamma=\frac{7}{5}=1+\frac{2}{f} \Rightarrow f=5\right)$ $C_{P}=\left(1+\frac{n}{2}\right) R$ $C_{P}=\frac{7}{2} R$ And molecule are rigid diatomic.
MHT-CET 2020
Kinetic Theory of Gases
139324
For a gas, $\frac{R}{C_{v}}=0.4$ where ' $R$ ' is universal gas constant and $C_{V}$ is the molar specific heat at constant volume. The gas is made up of molecule which are
1 Monoatomic
2 rigid diatomic
3 non-rigid diatomic
4 polyatomic
Explanation:
B Given that, $\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}=0.4$ $\mathrm{R}=0.4 \mathrm{C}_{\mathrm{V}}$ $\because \mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ So, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=0.4 \mathrm{C}_{\mathrm{V}}$ $\mathrm{C}_{\mathrm{P}}=1.4 \mathrm{C}_{\mathrm{V}}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=1.4=\frac{7}{5}$ Ratio of $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{7}{5}$, Therefore, gas is made up of rigid diatomic molecule.
MHT-CET 2020
Kinetic Theory of Gases
139325
The molar specific heats of an ideal gas at constant pressure and constant volume are denoted by $C_{p}$ and $C_{v}$ respectively. If $\gamma=\frac{C_{p}}{C_{v}}$ and $R$ is the universal gas constant, then $C_{p}$ is equal to
1 $\frac{\gamma-1}{\mathrm{R}}$
2 $\frac{(\gamma-1)^{2}}{\mathrm{R}}$
3 $\frac{\gamma-1}{\gamma \mathrm{R}}$
4 $\frac{\gamma \mathrm{R}}{\gamma-1}$
Explanation:
D Given that, $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma$ We know $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ From eq (i), $\frac{\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{V}}}=\gamma-1$ $\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}=\gamma-1 \quad\left(\because \mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}\right)$ $\frac{\gamma \mathrm{R}}{\mathrm{C}_{\mathrm{P}}}=\gamma-1 \quad\left(\because \mathrm{C}_{\mathrm{V}}=\frac{\mathrm{C}_{\mathrm{P}}}{\gamma}\right)$ $\mathrm{C}_{\mathrm{P}}=\frac{\gamma \mathrm{R}}{\gamma-1}$
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Kinetic Theory of Gases
139319
Specific heat of a gas undergoing adiabatic change is
1 Zero
2 Infinite
3 Positive
4 Negative
Explanation:
A Specific heat of a gas undergoing adiabatic change is zero because during adiabatic process no heat is supplied to system.
AP EAMCET-23.08.2021
Kinetic Theory of Gases
139323
For an ideal gas, if the ratio of Molar specific heats $\gamma=1.4$, then the specific heat at constant pressure $C_{P}$, specific heat at constant volume $C_{V}$ and corresponding molecule are respectively
C Ratio of molar specific heat $\gamma=1.4$ $\frac{C_{P}}{C_{V}}=\gamma=1.4$ $\frac{C_{P}}{C_{V}}=\frac{14}{10}=\frac{7}{5} \quad\left(\because \frac{C_{P}}{C_{V}}=\frac{7}{5}\right.$ so, diatomic rigid $)$ $C_{V}=\frac{f}{2} R$ So, $C_{V}=\frac{5}{2} R \quad\left(\because \gamma=\frac{7}{5}=1+\frac{2}{f} \Rightarrow f=5\right)$ $C_{P}=\left(1+\frac{n}{2}\right) R$ $C_{P}=\frac{7}{2} R$ And molecule are rigid diatomic.
MHT-CET 2020
Kinetic Theory of Gases
139324
For a gas, $\frac{R}{C_{v}}=0.4$ where ' $R$ ' is universal gas constant and $C_{V}$ is the molar specific heat at constant volume. The gas is made up of molecule which are
1 Monoatomic
2 rigid diatomic
3 non-rigid diatomic
4 polyatomic
Explanation:
B Given that, $\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}=0.4$ $\mathrm{R}=0.4 \mathrm{C}_{\mathrm{V}}$ $\because \mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ So, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=0.4 \mathrm{C}_{\mathrm{V}}$ $\mathrm{C}_{\mathrm{P}}=1.4 \mathrm{C}_{\mathrm{V}}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=1.4=\frac{7}{5}$ Ratio of $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{7}{5}$, Therefore, gas is made up of rigid diatomic molecule.
MHT-CET 2020
Kinetic Theory of Gases
139325
The molar specific heats of an ideal gas at constant pressure and constant volume are denoted by $C_{p}$ and $C_{v}$ respectively. If $\gamma=\frac{C_{p}}{C_{v}}$ and $R$ is the universal gas constant, then $C_{p}$ is equal to
1 $\frac{\gamma-1}{\mathrm{R}}$
2 $\frac{(\gamma-1)^{2}}{\mathrm{R}}$
3 $\frac{\gamma-1}{\gamma \mathrm{R}}$
4 $\frac{\gamma \mathrm{R}}{\gamma-1}$
Explanation:
D Given that, $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma$ We know $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ From eq (i), $\frac{\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{V}}}=\gamma-1$ $\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}=\gamma-1 \quad\left(\because \mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}\right)$ $\frac{\gamma \mathrm{R}}{\mathrm{C}_{\mathrm{P}}}=\gamma-1 \quad\left(\because \mathrm{C}_{\mathrm{V}}=\frac{\mathrm{C}_{\mathrm{P}}}{\gamma}\right)$ $\mathrm{C}_{\mathrm{P}}=\frac{\gamma \mathrm{R}}{\gamma-1}$
139319
Specific heat of a gas undergoing adiabatic change is
1 Zero
2 Infinite
3 Positive
4 Negative
Explanation:
A Specific heat of a gas undergoing adiabatic change is zero because during adiabatic process no heat is supplied to system.
AP EAMCET-23.08.2021
Kinetic Theory of Gases
139323
For an ideal gas, if the ratio of Molar specific heats $\gamma=1.4$, then the specific heat at constant pressure $C_{P}$, specific heat at constant volume $C_{V}$ and corresponding molecule are respectively
C Ratio of molar specific heat $\gamma=1.4$ $\frac{C_{P}}{C_{V}}=\gamma=1.4$ $\frac{C_{P}}{C_{V}}=\frac{14}{10}=\frac{7}{5} \quad\left(\because \frac{C_{P}}{C_{V}}=\frac{7}{5}\right.$ so, diatomic rigid $)$ $C_{V}=\frac{f}{2} R$ So, $C_{V}=\frac{5}{2} R \quad\left(\because \gamma=\frac{7}{5}=1+\frac{2}{f} \Rightarrow f=5\right)$ $C_{P}=\left(1+\frac{n}{2}\right) R$ $C_{P}=\frac{7}{2} R$ And molecule are rigid diatomic.
MHT-CET 2020
Kinetic Theory of Gases
139324
For a gas, $\frac{R}{C_{v}}=0.4$ where ' $R$ ' is universal gas constant and $C_{V}$ is the molar specific heat at constant volume. The gas is made up of molecule which are
1 Monoatomic
2 rigid diatomic
3 non-rigid diatomic
4 polyatomic
Explanation:
B Given that, $\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}=0.4$ $\mathrm{R}=0.4 \mathrm{C}_{\mathrm{V}}$ $\because \mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ So, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=0.4 \mathrm{C}_{\mathrm{V}}$ $\mathrm{C}_{\mathrm{P}}=1.4 \mathrm{C}_{\mathrm{V}}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=1.4=\frac{7}{5}$ Ratio of $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{7}{5}$, Therefore, gas is made up of rigid diatomic molecule.
MHT-CET 2020
Kinetic Theory of Gases
139325
The molar specific heats of an ideal gas at constant pressure and constant volume are denoted by $C_{p}$ and $C_{v}$ respectively. If $\gamma=\frac{C_{p}}{C_{v}}$ and $R$ is the universal gas constant, then $C_{p}$ is equal to
1 $\frac{\gamma-1}{\mathrm{R}}$
2 $\frac{(\gamma-1)^{2}}{\mathrm{R}}$
3 $\frac{\gamma-1}{\gamma \mathrm{R}}$
4 $\frac{\gamma \mathrm{R}}{\gamma-1}$
Explanation:
D Given that, $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma$ We know $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ From eq (i), $\frac{\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{V}}}=\gamma-1$ $\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}=\gamma-1 \quad\left(\because \mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}\right)$ $\frac{\gamma \mathrm{R}}{\mathrm{C}_{\mathrm{P}}}=\gamma-1 \quad\left(\because \mathrm{C}_{\mathrm{V}}=\frac{\mathrm{C}_{\mathrm{P}}}{\gamma}\right)$ $\mathrm{C}_{\mathrm{P}}=\frac{\gamma \mathrm{R}}{\gamma-1}$
139319
Specific heat of a gas undergoing adiabatic change is
1 Zero
2 Infinite
3 Positive
4 Negative
Explanation:
A Specific heat of a gas undergoing adiabatic change is zero because during adiabatic process no heat is supplied to system.
AP EAMCET-23.08.2021
Kinetic Theory of Gases
139323
For an ideal gas, if the ratio of Molar specific heats $\gamma=1.4$, then the specific heat at constant pressure $C_{P}$, specific heat at constant volume $C_{V}$ and corresponding molecule are respectively
C Ratio of molar specific heat $\gamma=1.4$ $\frac{C_{P}}{C_{V}}=\gamma=1.4$ $\frac{C_{P}}{C_{V}}=\frac{14}{10}=\frac{7}{5} \quad\left(\because \frac{C_{P}}{C_{V}}=\frac{7}{5}\right.$ so, diatomic rigid $)$ $C_{V}=\frac{f}{2} R$ So, $C_{V}=\frac{5}{2} R \quad\left(\because \gamma=\frac{7}{5}=1+\frac{2}{f} \Rightarrow f=5\right)$ $C_{P}=\left(1+\frac{n}{2}\right) R$ $C_{P}=\frac{7}{2} R$ And molecule are rigid diatomic.
MHT-CET 2020
Kinetic Theory of Gases
139324
For a gas, $\frac{R}{C_{v}}=0.4$ where ' $R$ ' is universal gas constant and $C_{V}$ is the molar specific heat at constant volume. The gas is made up of molecule which are
1 Monoatomic
2 rigid diatomic
3 non-rigid diatomic
4 polyatomic
Explanation:
B Given that, $\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}=0.4$ $\mathrm{R}=0.4 \mathrm{C}_{\mathrm{V}}$ $\because \mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ So, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=0.4 \mathrm{C}_{\mathrm{V}}$ $\mathrm{C}_{\mathrm{P}}=1.4 \mathrm{C}_{\mathrm{V}}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=1.4=\frac{7}{5}$ Ratio of $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{7}{5}$, Therefore, gas is made up of rigid diatomic molecule.
MHT-CET 2020
Kinetic Theory of Gases
139325
The molar specific heats of an ideal gas at constant pressure and constant volume are denoted by $C_{p}$ and $C_{v}$ respectively. If $\gamma=\frac{C_{p}}{C_{v}}$ and $R$ is the universal gas constant, then $C_{p}$ is equal to
1 $\frac{\gamma-1}{\mathrm{R}}$
2 $\frac{(\gamma-1)^{2}}{\mathrm{R}}$
3 $\frac{\gamma-1}{\gamma \mathrm{R}}$
4 $\frac{\gamma \mathrm{R}}{\gamma-1}$
Explanation:
D Given that, $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma$ We know $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ From eq (i), $\frac{\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{V}}}=\gamma-1$ $\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}=\gamma-1 \quad\left(\because \mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}\right)$ $\frac{\gamma \mathrm{R}}{\mathrm{C}_{\mathrm{P}}}=\gamma-1 \quad\left(\because \mathrm{C}_{\mathrm{V}}=\frac{\mathrm{C}_{\mathrm{P}}}{\gamma}\right)$ $\mathrm{C}_{\mathrm{P}}=\frac{\gamma \mathrm{R}}{\gamma-1}$