Question

A closed container constrains a homogeneous mixture made of, 2 moles of an ideal monatomic gas($\gamma=\frac{5}{3}$) and one mole of an ideal diatomic gas($\gamma=\frac{7}{5}$). Here, γ is the ratio of the specific heat at content pressure and constant volume of an ideal gas. The gas mixture does a work of 66 joules when heated at constant pressure. The change in its internal energy is ____Joule.

• A. -187J
• B. 121J
• C. -121J
• D. 187J

Answer 1

Answer: (B)

121J

Answer 2

The correct option is (B): 121 J.
For the mixture of the gas,

$\gamma=\frac{n_1Cp_1+n_2Cp_2}{n_1Cv_1+n_2Cv_2}=\frac{2\times\frac{5}{2}R+1\times\frac{7}{2}R}{2\times\frac{3}{2}R+1\times\frac{5}{2}R}=\frac{17}{11}$

as gas in heated constant pressure,

$W=nR\Delta T$

$\Delta U=nC_v\Delta T$

$Q=nC_p\Delta T$

Now, $\frac{\Delta U}{W}=\frac{\Delta U}{Q-\Delta U}=\frac{1}{\frac{Q}{\Delta U}-1}$

$\frac{\Delta U}{W}=\frac{1}{\frac{C_P}{C_V}-1}=\frac{1}{\gamma-1}\,\,;\,\,\frac{\Delta U}{W}=\frac{1}{\frac{17}{11}-1}\,\,;\,\,\frac{\Delta U}{66}=\frac{11}{6}\,\,;\,\,\Delta U=121J$