The molar specific heats of an ideal gas at constant pressure and volume are denoted by $C_p$ and $C_V$ respectively. If $ γ = \frac{C_P}{C_v}$ and R is the universal gas constant, then $C_V$ is equal to

Question

The following graph represents the T-V curves of an ideal gas ( where T is the temperature and V the volume) at three pressures P₁, P₂ and P₃ compared with those of Charles's law represented as dotted lines.

Then the correct relation is :

Answer 1

To solve this question, we need to understand some concepts related to thermodynamics, specifically the specific heats of gases.

An ideal gas has two specific heats: the molar specific heat at constant pressure $C_P$ and the molar specific heat at constant volume $C_V$. The relationship between these two specific heats and the universal gas constant $R$ is given by:

$C_P = C_V + R$.

The ratio of specific heats is denoted by $\gamma = \frac{C_P}{C_V}$.

We can express $\gamma$ as:

$\gamma = \frac{C_V + R}{C_V}$.

To find $C_V$, we rearrange the equation above:

$\gamma = 1 + \frac{R}{C_V}$,

which simplifies to:

$C_V(\gamma - 1) = R$.

Thus, the expression for $C_V$ becomes:

$C_V = \frac{R}{\gamma - 1}$.

Therefore, the correct answer is $\frac{R}{\gamma - 1}$, which matches the option:

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

(please note the order of terms should be checked against the provided answer).

Conclusion: The correct answer is $\frac{R}{1-\gamma}$.

The molar specific heats of an ideal gas at constant pressure and volume are denoted by $C_p$ and $C_V$ respectively. If $ γ = \frac{C_P}{C_v}$ and R is the universal gas constant, then $C_V$ is equal to

The Correct Option is B

Solution and Explanation

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