/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 78 The work of \(146 \mathrm{~kJ}\)... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 78

The work of \(146 \mathrm{~kJ}\) is performed in order to compress 1 kilo mole of a gas adiabatically, and in this process, the temperature of the gas increases by \(7^{\circ} \mathrm{C}\). The gas is \(\left(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\) (A) diatomic. (B) triatomic. (C) a mixture of monoatomic and diatomic. (D) monatomic.

Short Answer

Expert verified
Based on the analysis, it can be concluded that the given gas is a mixture of monoatomic and diatomic gases (option C).

Step by step solution

01

Calculate the temperature change in Kelvin

We are given that the temperature increases by 7掳C. To convert this temperature change to Kelvin, we need to remember that the Kelvin scale has the same increments as the Celsius scale: \(\Delta T = 7 K\)
02

Use the adiabatic work formula to calculate the heat capacity at constant volume

We are given that the work done is 146 kJ, which we need to convert to J: \(W = 146 脳 10^3 J\) We also know that there is 1 kilomole of the gas involved in the process, which means \(n = 1\). Since \(W = nC_V\Delta T\), we can rewrite this formula to find the heat capacity at constant volume: \(C_V = \frac{W}{n\Delta T}\) Now we can substitute the values: \(C_V = \frac{146 脳 10^3}{1 脳 7}\) \(C_V \approx 20857 J\,K^{-1}\, mol^{-1}\)
03

Compare the obtained value with the heat capacity values for different types of gases

We have the following heat capacity at constant volume for different gases: - Monatomic: \(C_V = \frac{3}{2}R\) - Diatomic: \(C_V = \frac{5}{2}R\) - Triatomic: \(C_V = \frac{f}{2}R\), where \(f\) is the degrees of freedom (usually 3). Substitute the given value of R: - Monatomic: \(C_V = \frac{3}{2} 脳 8.3 \approx 12.45 J\,K^{-1}\, mol^{-1}\) - Diatomic: \(C_V = \frac{5}{2} 脳 8.3 \approx 20.75 J\,K^{-1}\, mol^{-1}\) - Triatomic: \(C_V = \frac{3}{2} 脳 8.3 = 12.45 J\,K^{-1}\, mol^{-1}\) (assuming linear triatomic gas with 3 degrees of freedom) As we can see, the obtained value of \(C_V \approx 20857 J\,K^{-1}\, mol^{-1}\) does not match any of these values. This indicates that the given gas is a mixture.
04

Conclusion

Based on the analysis, it can be concluded that the given gas is a mixture of monoatomic and diatomic gases (option C).

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Most popular questions from this chapter

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