Problem 96 - A Complete Resource Book in Physics for JEE Main

Question

If three moles of monatomic gas is mixed with 1 moles diatomic gas, the resultant value of \(\gamma\) for the mixture is (A) \(1.66\) (B) \(1.50\) (C) \(1.40\) (D) \(1.57\)

Step-by-Step Solution

Answer

The effective adiabatic index for the mixture is \(\gamma = \frac{11}{7} \approx 1.57\). Therefore, the correct answer is (D) \(1.57\).

Step 1: Recall the adiabatic index formula#

In thermodynamics, the adiabatic index (\(\gamma\)) is the ratio of the molar heat capacity at constant pressure (\(C_p\)) to the molar heat capacity at constant volume (\(C_v\)). Mathematically, it is expressed as follows: \[\gamma = \frac{C_p}{C_v}\]

Step 2: Determine the individual heat capacities and adiabatic indexes#

For monatomic and diatomic gases, the molar heat capacities are known: - For monatomic gases, \(C_v = \frac{3}{2}R\) and \(C_p = \frac{5}{2}R\), where R is the gas constant. Therefore, their adiabatic index (let's call it \(\gamma_1\)) is \[\gamma_1 = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3}\] - For diatomic gases, \(C_v = \frac{5}{2}R\) and \(C_p = \frac{7}{2}R\). Therefore, their adiabatic index (let's call it \(\gamma_2\)) is \[\gamma_2 = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5}\]

Step 3: Calculate the total volumes and heat capacities of the mixture#

First, let's find the total volume of the system (V). Since both gases are ideal, their volumes are additive. With 3 moles of monatomic gas and 1 mole of diatomic gas, the total number of moles of the mixture (n) is \[n = 3 + 1 = 4\] Thus, the volume fraction for monatomic and diatomic gases are \[\frac{V_1}{V} = \frac{3}{4}\] and \[\frac{V_2}{V} = \frac{1}{4}\] Now we can find the total heat capacities of the system at constant volume (Cv) and at constant pressure (Cp) as follows: \[C_v = \frac{3}{4} \cdot \frac{3}{2}R + \frac{1}{4} \cdot \frac{5}{2}R\] \[C_p = \frac{3}{4} \cdot \frac{5}{2}R + \frac{1}{4} \cdot \frac{7}{2}R\]

Step 4: Calculate the effective adiabatic index#

Now that we have the total heat capacities (Cv and Cp), we can calculate the adiabatic index for the mixture using the initial formula: \[\gamma = \frac{C_p}{C_v}\] Substitute the expressions for Cp and Cv as calculated in the previous step: \[\gamma = \frac{\frac{3}{4} \cdot \frac{5}{2}R + \frac{1}{4} \cdot \frac{7}{2}R}{\frac{3}{4} \cdot \frac{3}{2}R + \frac{1}{4} \cdot \frac{5}{2}R}\] Cancel the R terms and simplify: \[\gamma = \frac{15 + 7}{9 + 5} = \frac{22}{14} = \frac{11}{7}\]

Step 5: Determine the answer choice#

Finally, to find the answer, approximate the numeric value of \(\gamma\): \[\gamma = \frac{11}{7} \approx 1.57\] Thus, the correct answer is: (D) \(1.57\).