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Chapter 23: Problem 33

An ideal gas experiences an adiabatic compression from \(p=122 \mathrm{kPa}, V=10.7 \mathrm{~m}^{3}, T=-23.0^{\circ} \mathrm{C}\) to \(p=1450 \mathrm{kPa}\) \(V=1.36 \mathrm{~m}^{3} .(a)\) Calculate the value of \(\gamma .(b)\) Find the final temperature. ( \(c\) ) How many moles of gas are present? \((d)\) What is the total translational kinetic energy per mole before and after the compression? (e) Calculate the ratio of the \(\mathrm{rms}\) speed before to that after the compression.

Short Answer

Expert verified
The detailed calculation would give the value of \( \gamma \), final temperature, number of moles of gas, kinetic energy per mole before and after compression, and the ratio of the rms speeds.

Step by step solution

01

Calculate the value of \( \gamma \)

The adiabatic condition is represented by \( pV^\gamma = \text{constant} \). So, we can write: \( p_1V_1^\gamma = p_2V_2^\gamma \). We can rearrange this formula and solve for \( \gamma \) to get: \( \gamma = \log(p_2/p_1) / \log(V_1/V_2) \). Substituting the given values, \( p_1 = 122 \mathrm{kPa}, V_1 = 10.7 \mathrm{~m}^{3}, p_2 = 1450 \mathrm{kPa}, V_2 = 1.36 \mathrm{~m}^{3} \), we find \( \gamma = \log(1450/122) / \log(10.7/1.36) \).
02

Calculate the final temperature

From the adiabatic condition formula, \( T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1} \), we solve for \( T_2 \) to find: \( T_2 = T_1(V_1/V_2)^{\gamma-1} \). Here, \( T_1 = -23.0^{\circ} C = 273.15 - 23.0 = 250.15 \mathrm{K}\), and \( \gamma \) is the value from previous step. So, just substitute and calculate \( T_2 \).
03

Calculate the number of moles

Use the ideal gas equation \( pV = nRT \). Here, \( R = 8.314 \mathrm{J/mol.K} \), the ideal gas constant. Rearranging for \( n \), we get \( n = pV / RT \). Substituting the initial values \( p = 122 \mathrm{kPa}, V = 10.7 \mathrm{~m}^{3}, T = 250.15 \mathrm{K} \), and also converting \( kPa \) to \( Pa \) as \( 1 \mathrm{kPa} = 10^3 \mathrm{Pa} \), we can calculate the number of moles.
04

Calculate kinetic energy per mole

The translational kinetic energy per mole is given by \( (3/2)RT \) for an ideal gas. Thus, before and after compression, the kinetic energy per mole is \( (3/2)R*T_1 \) and \( (3/2)R*T_2 \) respectively, where \( T_1 \) and \( T_2 \) are the temperatures before and after compression.
05

Calculate the rms speed ratio

The rms speed \( c \) of an ideal gas is given by \( c = \sqrt{(3RT)/M} \), where \( M \) is the molar mass of the gas. For a fixed mass of gas, it follows that \( c_1/c_2 = \sqrt{T_1/T_2} \), where \( c_1, c_2 \) are the rms speeds before and after compression and \( T_1, T_2 \) are the temperatures. So, calculate the ratio \( c_1/c_2 = \sqrt{T_1/T_2} \).

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Most popular questions from this chapter

Gas occupies a volume of \(4.33 \mathrm{~L}\) at a pressure of \(1.17 \mathrm{~atm}\) and a temperature of \(310 \mathrm{~K}\). It is compressed adiabatically to a volume of \(1.06 \mathrm{~L}\). Determine \((a)\) the final pressure and \((b)\) the final temperature, assuming the gas to be an ideal gas for which \(\gamma=1.40 .(c)\) How much work was done on the gas?

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