Question 780: Thermodynamics & KTG

Question

If $C_p$ and $C_v$ are molar specific heats of an ideal gas at constant pressure and volume respectively and ' $\gamma$ ' is $\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}$ then $\mathrm{C}_{\mathrm{p}}=$
( $\mathrm{R}=$ universal gas constant)

Accepted Answer

To understand how to derive the expression for $ C_p $, the molar specific heat of an ideal gas at constant pressure, let's go through the following steps:
We start with the given relation for the heat capacity ratio (also known as the adiabatic index), $\gamma$, which is defined as:
$    \gamma = \frac{C_p}{C_v}    $
This implies:
$    \frac{C_v}{C_p} = \frac{1}{\gamma}    $
We can express the difference in specific heats as:
$    \frac{C_v - C_p}{C_p} = \frac{1 - \gamma}{\gamma}    $
Rearranging the terms gives:
$    \frac{-(C_p - C_v)}{C_p} = \frac{1 - \gamma}{\gamma}    $
Since the relation between the specific heats is $ C_p - C_v = R $ (where $ R $ is the universal gas constant), we substitute this into the equation:
$    \frac{-R}{C_p} = \frac{1 - \gamma}{\gamma}    $
Simplifying, we have:
$    \frac{R}{C_p} = \frac{\gamma - 1}{\gamma}    $
Finally, solving for $ C_p $, we find:
$    C_p = \frac{\gamma R}{\gamma - 1}    $
Thus, the molar specific heat at constant pressure $ C_p $ for an ideal gas is given by the expression $\frac{\gamma R}{\gamma - 1}$.

Answer

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

Answer

$\gamma \mathrm{R}$

Answer

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

Answer

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

Thermodynamics & KTG

If C_p and C_v are molar specific heats of an ideal gas at constant pressure and volume respectively and ' \gamma ' is \mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}} then \mathrm{C}_{\mathrm{p}}= ( \mathrm{R}= universal gas constant)

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