Question 960: Thermodynamics & KTG

Question

If ' $C_P$ ' and ' $C_V$ ' are molar specific heats of an ideal gas at constant pressure and volume respectively. If ' $\lambda$ ' is the ratio of two specific heats and ' $R$ ' is universal gas constant then ' $C_p$ ' is equal to

Accepted Answer

Given that the ratio of the molar specific heats at constant pressure ($C_p$) and constant volume ($C_v$) is $\gamma$, we need to find the expression for $C_p$. The universal gas constant is denoted by $R$.
The relationship between $C_p$ and $C_v$ for an ideal gas is given by Mayer's relation:
$ C_p - C_v = R $
By substituting $C_p$ from the above relation into the expression for $\gamma$, we have:
$ \gamma = \frac{C_p}{C_v} = \frac{C_p}{C_p - R} $
To solve for $C_p$, multiply both sides by $(C_p - R)$:
$ \gamma (C_p - R) = C_p $
Expanding and rearranging terms, we get:
$ \gamma C_p - C_p = \gamma R $
Factor out $C_p$ on the left side:
$ C_p (\gamma - 1) = \gamma R $
Finally, solve for $C_p$:
$ C_p = \frac{\gamma R}{\gamma - 1} $
Thus, $C_p$ is equal to $\frac{\gamma R}{\gamma - 1}$.

Answer

$\frac{R \gamma}{\gamma-1}$

Answer

$\gamma R$

Answer

$\frac{1+\gamma}{1-\gamma}$

Answer

$\frac{R}{\gamma-1}$

Thermodynamics & KTG

If ' C_P ' and ' C_V ' are molar specific heats of an ideal gas at constant pressure and volume respectively. If ' \lambda ' is the ratio of two specific heats and ' R ' is universal gas constant then ' C_p ' is equal to

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