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Chapter 8: Problem 1164

A diatomic gas initially at \(18^{\circ} \mathrm{C}\) is Compressed adiabatically to one eighth of its original volume. The temperature after Compression will be (A) \(10^{\circ} \mathrm{C}\) (B) \(668 \mathrm{~K}\) (C) \(887^{\circ} \mathrm{C}\) (D) \(144^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The final temperature after adiabatic compression of the diatomic gas is \(T_f = 668^{\circ}\mathrm{C}\) or \(941.15\mathrm{K}\). The correct answer is (B) 668 K.

Step by step solution

01

Write down initial and final conditions given by the problem.

The initial temperature of the gas is \(18^{\circ}\mathrm{C}\), and we are told it is compressed adiabatically to 1/8 of its original volume. Therefore, we have: - Initial volume: \(V_i\) - Initial temperature: \(T_i = 18^{\circ}\mathrm{C}\) - Final volume: \(V_f = \frac{1}{8} V_i\) And we want to find the final temperature, \(T_f\).
02

Convert initial temperature to Kelvin.

In order to work with the ideal gas law and adiabatic equations, we need to convert the initial temperature to Kelvin: \(T_i = 18^{\circ}\mathrm{C}+ 273.15 = 291.15 \mathrm{K}\)
03

Use adiabatic process equation for diatomic gas.

The adiabatic process equation for the diatomic gas is: \(P_i V_i^{\gamma} = P_f V_f^{\gamma}\) We will also use the ideal gas law: \(PV = nRT\), where we can assume the number of moles (n) and the gas constant (R) are constant in this problem.
04

Substitute ideal gas law into adiabatic process equation.

Since we are only interested in temperature, we can substitute the ideal gas law equation for pressure into the adiabatic process equation: \(\frac{nR T_i}{V_i}\) \(V_i^{\gamma} = \frac{nRT_f}{V_f} V_f^{\gamma}\)
05

Solve the equation for the final temperature, \(T_f\).

By canceling out the constants and simplifying the equation, we get: \(T_i V_i^{\gamma - 1} = T_f V_f^{\gamma -1}\) Now solving for the final temperature, \(T_f\): \(T_f = T_i \left(\frac{V_i}{V_f}\right)^{\gamma -1} = 291.15 \mathrm{K}\left(\frac{V_i}{\frac{1}{8} V_i}\right)^{7/5 -1}\) \(T_f = 291.15 \mathrm{K}\left(8\right)^{2/5}\)
06

Calculate final temperature and convert to Celsius.

Now we calculate the final temperature in Kelvin and then convert it back to Celsius: \(T_f = 291.15 \mathrm{K}\times \left(8\right)^{2/5} = 941.15 \mathrm{K}\) \(T_f = 941.15 \mathrm{K} - 273.15 = \boxed{668^{\circ} \mathrm{C}}\) Thus, the correct answer is (B) 668 K.

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