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

The Correct Option is B

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, \(\frac{\Delta U}{W}=\frac{\Delta U}{Q-\Delta U}=\frac{1}{\frac{Q}{\Delta U}-1}\)

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

Answer 2

First, we find the heat capacities at constant volume for monatomic and diatomic gases.

For a monatomic ideal gas, the heat capacity at constant volume (C1) is 3R. This is because a monatomic gas has three degrees of freedom (motion in x, y, and z directions).

For a diatomic ideal gas, the heat capacity at constant volume (C2) is 5R. This is because a diatomic gas has five degrees of freedom (motion in x, y, and z directions, and rotation about two axes).

Next, we calculate the average heat capacity at constant volume for the gas mixture (Comix). This is a weighted average based on the number of moles of each gas:

\(\text{Comix} = \frac{n_1 \cdot C_1 + n_2 \cdot C_2}{n_1 + n_2} = \frac{2 \cdot 3R + 1 \cdot 5R}{2 + 1} = \frac{11R}{3} = \frac{11R}{6}\)

The change in internal energy (ΔU) for a given number of moles (n) and change in temperature (ΔT) is:

\(\Delta U = n \cdot C \cdot \Delta T\)

Given that the work done by the system at constant pressure (W) is:

\(𝑊=𝑛⋅𝑅⋅Δ𝑇\)

We can replace ΔT in the equation for ΔU:

\(\Delta U = n \cdot \text{Comix} \cdot \Delta T\)
 \(W = n \cdot R \cdot \Delta T\)

So,
\(\Delta U = \text{Comix} \cdot \frac{W}{R}\)

Substitute the values:
\(\Delta U = \left( \frac{11R}{6} \cdot \frac{W}{R} \right)\)
\(\Delta U = \frac{11W}{6}\)

If W is 66 Joules, then:

\(\Delta U = \frac{11 \cdot 66}{6} = 121 \, \text{Joules}\)

So, the change in internal energy is 121 Joules.