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NEET CHEMISTRYThermodynamicsEasy

Question

Equal volumes of two monoatomic gases, A and B, at same temperature and pressure are mixed. The ratio of specific heats (Cp/CvC_p/C_v) of the mixture will be:

A

1.501.50

B

3.33.3

C

1.671.67

D

0.830.83

Step-by-Step Solution

Since equal volumes of gases are taken at the same temperature and pressure, their number of moles will be equal, let it be nn. For monoatomic gases, Cv=32RC_v = \frac{3}{2}R and Cp=52RC_p = \frac{5}{2}R. The molar heat capacity at constant volume for the mixture is given by: Cv(mix)=n1Cv1+n2Cv2n1+n2=n(32R)+n(32R)n+n=32RC_{v(\text{mix})} = \frac{n_1C_{v1} + n_2C_{v2}}{n_1 + n_2} = \frac{n(\frac{3}{2}R) + n(\frac{3}{2}R)}{n + n} = \frac{3}{2}R The molar heat capacity at constant pressure for the mixture is given by: Cp(mix)=n1Cp1+n2Cp2n1+n2=n(52R)+n(52R)n+n=52RC_{p(\text{mix})} = \frac{n_1C_{p1} + n_2C_{p2}}{n_1 + n_2} = \frac{n(\frac{5}{2}R) + n(\frac{5}{2}R)}{n + n} = \frac{5}{2}R Therefore, the ratio of specific heats for the mixture is: γmix=Cp(mix)Cv(mix)=52R32R=53≈1.67\gamma_{\text{mix}} = \frac{C_{p(\text{mix})}}{C_{v(\text{mix})}} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3} \approx 1.67 Alternatively, since both are monoatomic gases, the mixture will also behave as a monoatomic gas, so γ=53≈1.67\gamma = \frac{5}{3} \approx 1.67.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicsvolumesmonoatomictemperaturepressurespecific

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