Question 718: Thermodynamics & KTG

Question

The molar specific heat of an ideal gas at constant pressure and constant volume is ' $\mathrm{C}_{\mathrm{p}}$ ' and ' $\mathrm{C}_{\mathrm{v}}$ ' respectively. If ' R ' is a universal gas constant and the ratio of ' $\mathrm{C}_{\mathrm{p}}$ ' to ' $\mathrm{C}_{\mathrm{v}}$ ' is $\gamma$, then ' $\mathrm{C}_{\mathrm{p}}$ ' is equal to

Accepted Answer

Given:
Molar specific heat at constant pressure $= C_p$
Molar specific heat at constant volume $= C_v$
Universal gas constant $= R$
Ratio $\displaystyle \gamma = \frac{C_p}{C_v} $
Standard Relations:
Relation between $C_p, C_v$ and $R$:
$$    C_p - C_v = R     $$
Given:
$$    \gamma = \frac{C_p}{C_v}    $$
Step 1: Express $C_p$ in terms of $C_v$:
$$ C_p = \gamma C_v $$
Step 2: Substitute $C_p$ in the first equation:
$$ C_p - C_v = R \ \Rightarrow (\gamma C_v) - C_v = R \ \Rightarrow (\gamma - 1)C_v = R  $$
Step 3: Express $C_v$:
$$ C_v = \frac{R}{\gamma - 1} $$
Step 4: Express $C_p$ in terms of $R$ and $\gamma$:
Recall, $C_p = \gamma C_v$, so:
$$ C_p = \gamma \cdot \frac{R}{\gamma - 1} = \frac{\gamma R}{\gamma - 1} $$
Final Answer:
The correct expression is:
$$ C_p = \frac{\gamma R}{\gamma - 1} $$
Option C
is correct.

Answer

$\left(\frac{\gamma-1}{\gamma+1}\right) \mathrm{R}$

Answer

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

Answer

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

Answer

$\frac{\mathrm{R} \gamma}{(\gamma+1)}$

Thermodynamics & KTG

Choose the correct answer: