HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q11 Solution: Two ideal gases have the same value of . What will be the value …

QPages 77-80: Q11: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q11 Solution: Two ideal gases have the same value of . What will be the value …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often take the mole‑fraction weighted average of the individual $gamma$ values, but $gamma$ is a ratio of heat capacities, not a linear property. Compute $C_{v, ext{mix}}=rac{n_1C_{v1}+n_2C_{v2}}{n_1+n_2}$ and $C_{p, ext{mix}}=rac{n_1C_{p1}+n_2C_{p2}}{n_1+n_2}$ first, then form $gamma_{ ext{mix}}=C_{p, ext{mix}}/C_{v, ext{mix}}$.

Two ideal gases have the same value of

C_p / C_v = \gamma

. What will be the value of this ratio for a mixture of the two gases in the ratio 1 : 2?

Step-by-Step Verified Solution

**Answer:**

\gamma_{mixture} = \gamma

. **Step-by-Step Explanation:** For both gases,

C_v = \frac{R}{\gamma-1}

and

C_p = \frac{\gamma R}{\gamma-1}

. For mixture of

n_1

and

n_2

moles:

C_v = \frac{n_1C_{v1} + n_2C_{v2}}{n_1+n_2} = \frac{(n_1+n_2)\frac{R}{\gamma-1}}{n_1+n_2} = \frac{R}{\gamma-1}

Similarly,

C_p = \frac{\gamma R}{\gamma-1}

. Hence

\gamma_{mixture} = \frac{C_p}{C_v} = \gamma

[citation:12].

QPages 77-80: Q11: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q11 Solution: Two ideal gases have the same value of . What will be the value …

Step-by-Step Verified Solution

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