Verified Solution • 2025–26 Session moderate

QPages 77-80: Q12: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q12 Solution: A mixture contains 1 mole of helium ) and 1 mole of hydrogen ). …

A mixture contains 1 mole of helium Cp=2.5R,Cv=1.5RC_p = 2.5 R, C_v = 1.5 R) and 1 mole of hydrogen Cp=3.5R,Cv=2.5RC_p = 3.5 R, C_v = 2.5 R). Calculate the values of CpC_p, CvC_v and γ\gamma for the mixture.

Step-by-Step Verified Solution

**Answer:** Cp=3RC_p = 3R, Cv=2RC_v = 2R, γ=1.5\gamma = 1.5. **Step-by-Step Explanation:** Given: 1 mole He (Cp=2.5RC_p = 2.5R, Cv=1.5RC_v = 1.5R) and 1 mole H_ (Cp=3.5RC_p = 3.5R, Cv=2.5RC_v = 2.5R). Mixture Cv=1×1.5R+1×2.5R2=4R2=2RC_v = \frac{1 \times 1.5R + 1 \times 2.5R}{2} = \frac{4R}{2} = 2R. Cp=Cv+R=3RC_p = C_v + R = 3R. γ=CpCv=3R2R=1.5\gamma = \frac{C_p}{C_v} = \frac{3R}{2R} = 1.5 [citation:12].

All Questions in this Chapter

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