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Chapter 8: Problem 1122

The work of \(62.25 \mathrm{KJ}\) is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by \(5^{\circ} \mathrm{C}\) The gas is $\mathrm{R}=8.3\\{\mathrm{~J} /(\mathrm{mole})\\}$ (A) triatomic (B) diatomic (C) monoatomic (D) a mixture of monoatomic and diatomic

Short Answer

Expert verified
The gas is a mixture of monoatomic and diatomic gases, as the calculated value of \(\gamma \approx 1.003\) is not close to the values for monoatomic (\(1.67\)), diatomic (\(1.4\)), or triatomic (\(1.33\)) gases. The answer is (D) a mixture of monoatomic and diatomic.

Step by step solution

01

Write down the adiabatic process equation

The equation for an adiabatic process is given by \(W = \frac{n R \Delta T}{\gamma - 1}\), where \(W\) is the work done, \(n\) is the number of moles, \(R\) is the gas constant, \(\Delta T\) is the change in temperature, and \(\gamma\) is the ratio of specific heat capacities.
02

Plug in the given values

We are given \(W = 62.25 \ \mathrm{KJ}\), \(n = 1 \ \mathrm{mole}\), \(R = 8.3 \ \mathrm{J/mole}\), and \(\Delta T = 5^{\circ} \mathrm{C} = 5 \ \mathrm{K}\). Plug these values into the adiabatic process equation: \(\frac{62.25 \times 10^3 \ \mathrm{J}}{1 \ \mathrm{mole}} = \frac{1 \ \mathrm{mole}\cdot 8.3 \ \mathrm{J/mole}\cdot 5 \ \mathrm{K}}{\gamma - 1}\)
03

Solve for γ

Simplify and solve the equation for γ: \(\frac{62.25 \times 10^3 \ \mathrm{J}}{1 \ \mathrm{mole}} = \frac{41.5 \ \mathrm{J}\cdot 5 \ \mathrm{K}}{\gamma - 1}\) \(\frac{62.25 \times 10^3 \ \mathrm{J}}{1 \ \mathrm{mole}} = \frac{207.5 \ \mathrm{J}\cdot \mathrm{K}}{\gamma - 1}\) \((\gamma - 1) = \frac{207.5 \ \mathrm{J}\cdot \mathrm{K}}{62.25 \times 10^3 \ \mathrm{J} / 1 \ \mathrm{mole}}\) \(\gamma = 1 + \frac{207.5 \ \mathrm{J} \cdot \mathrm{K}}{62.25 \times 10^3 \ \mathrm{J} / 1 \ \mathrm{mole}}\) \(\gamma \approx 1 + \frac{207.5}{62.25 \times 10^3}\) \(\gamma \approx 1.003\)
04

Determine the type of gas

Based on the calculated value of γ, we can now determine the type of gas: - For a monoatomic gas, γ = 5/3 = 1.67 - For a diatomic gas, γ = 7/5 = 1.4 - For a triatomic gas, γ = 4/3 = 1.33 The calculated value of γ (1.003) is not close to any of these values for monoatomic, diatomic, or triatomic gases. This indicates that the gas is likely a mixture of monoatomic and diatomic gases. Therefore, the answer is (D) a mixture of monoatomic and diatomic.

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