Class 11PhysicsKinetic Theory of GasesMolar Specific Heat of gas and relation between them (Mayer's formula)
MediumGiven below are observations on molar specific heats at room temperature of some common gases.
Gas Molar specific heat $(C_v)$ $(cal\, mol^{-1}\, K^{-1})$ Hydrogen $4.87$ Nitrogen $4.97$ Oxygen $5.02$ Nitric oxide $4.99$ Carbon monoxide $5.01$ Chlorine $6.17$
The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically,molar specific heat of a monatomic gas is $2.92 \; cal/mol\; K$. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?
| Gas | Molar specific heat $(C_v)$ $(cal\, mol^{-1}\, K^{-1})$ |
|---|---|
| Hydrogen | $4.87$ |
| Nitrogen | $4.97$ |
| Oxygen | $5.02$ |
| Nitric oxide | $4.99$ |
| Carbon monoxide | $5.01$ |
| Chlorine | $6.17$ |
The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically,molar specific heat of a monatomic gas is $2.92 \; cal/mol\; K$. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?
(N/A) The gases listed in the table are diatomic. Besides the translational degrees of freedom,they possess rotational degrees of freedom.
Heat supplied to the gas increases the average energy of all these modes of motion. Consequently,the molar specific heat of diatomic gases is higher than that of monatomic gases.
If only translational and rotational modes are considered,the theoretical molar specific heat of a diatomic gas is given by $C_v = \frac{5}{2} R$.
Using $R \approx 1.98 \; cal\, mol^{-1}\, K^{-1}$,we get $C_v = 2.5 \times 1.98 = 4.95 \; cal\, mol^{-1}\, K^{-1}$.
Most gases in the table show values close to $4.95 \; cal\, mol^{-1}\, K^{-1}$. However,chlorine has a significantly higher value $(6.17 \; cal\, mol^{-1}\, K^{-1})$. This indicates that at room temperature,chlorine molecules also possess vibrational degrees of freedom in addition to translational and rotational modes,which contribute to the total internal energy and thus increase the molar specific heat.
Heat supplied to the gas increases the average energy of all these modes of motion. Consequently,the molar specific heat of diatomic gases is higher than that of monatomic gases.
If only translational and rotational modes are considered,the theoretical molar specific heat of a diatomic gas is given by $C_v = \frac{5}{2} R$.
Using $R \approx 1.98 \; cal\, mol^{-1}\, K^{-1}$,we get $C_v = 2.5 \times 1.98 = 4.95 \; cal\, mol^{-1}\, K^{-1}$.
Most gases in the table show values close to $4.95 \; cal\, mol^{-1}\, K^{-1}$. However,chlorine has a significantly higher value $(6.17 \; cal\, mol^{-1}\, K^{-1})$. This indicates that at room temperature,chlorine molecules also possess vibrational degrees of freedom in addition to translational and rotational modes,which contribute to the total internal energy and thus increase the molar specific heat.
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Medium
View SolutionThe molar specific heat of an ideal gas at constant pressure and constant volume is $C_p$ and $C_v$ respectively. If $R$ is the universal gas constant and the ratio of $C_p$ to $C_v$ is $\gamma$,then $C_v=$
$Assertion:$ At a given temperature,the specific heat of a gas at constant pressure $(C_p)$ is always greater than its specific heat at constant volume $(C_v)$.
$Reason:$ When a gas is heated at constant volume,some extra heat is needed compared to that at constant pressure for doing work in expansion.
$Reason:$ When a gas is heated at constant volume,some extra heat is needed compared to that at constant pressure for doing work in expansion.
For all gases,$C_P - C_V$ is a constant. Does this imply that for all gases,$\frac{C_P}{C_V}$ is also a constant?
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