Question:

Two moles a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is :

Updated On: Nov 14, 2024
  • \(\frac{9}{4}R\)
  • \(\frac{7}{4}R\)
  • \(\frac{3}{2}R\)
  • \(\frac{5}{2}R\)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

For a monatomic gas, the molar specific heat at constant volume is \(C_V = \frac{3}{2}R\), and for a diatomic gas, \(C_V = \frac{5}{2}R\). The total heat capacity \(C_V\) of the mixture is given by the weighted average formula:

\[ C_V = \frac{n_1C_{V1} + n_2C_{V2}}{n_1 + n_2} \]

Where:

  • \(n_1 = 2\) is the number of moles of the monatomic gas.
  • \(n_2 = 6\) is the number of moles of the diatomic gas.
  • \(C_{V1} = \frac{3}{2}R\) is the molar specific heat of the monatomic gas.
  • \(C_{V2} = \frac{5}{2}R\) is the molar specific heat of the diatomic gas.

Substituting the values into the equation:

\[ C_V = \frac{2 \times \frac{3}{2}R + 6 \times \frac{5}{2}R}{2 + 6} = \frac{3R + 15R}{8} = \frac{18R}{8} = \frac{9}{4}R \]

Thus, the molar specific heat of the mixture at constant volume is \(\frac{9}{4}R\).

Was this answer helpful?
0
0

Top Questions on specific heat capacity

View More Questions

Questions Asked in JEE Main exam

View More Questions