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Chapter 13: Problem 41

A cylinder in a car engine takes \(V_{i}=4.50 \times 10^{-2} \mathrm{m}^{3}\) of air into the chamber at \(30^{\circ} \mathrm{C}\) and at atmospheric pressure. The piston then compresses the air to one-ninth of the original volume \(\left(0.111 \mathrm{V}_{\mathrm{i}}\right)\) and to 20.0 times the original pressure \(\left(20.0 P_{\mathrm{i}}\right) .\) What is the new temperature of the air?

Short Answer

Expert verified
Answer: The final temperature of the air after compression is approximately 672.27 K.

Step by step solution

01

Write down the initial and final conditions

We are given the following information: - Initial volume: \(V_i = 4.50 \times 10^{-2} \mathrm{m}^3\) - Initial temperature: \(T_i = 30^\circ\mathrm{C}\) - Initial pressure: \(P_i\) (atmospheric pressure) - Final volume: \(V_f = 0.111 V_i\) - Final pressure: \(P_f = 20.0 P_i\) Our goal is to find the final temperature \(T_f\).
02

Convert the temperatures to Kelvin

In order to apply the ideal gas law, we need to convert the temperatures from Celsius to Kelvin. $$ T_i = 30^\circ\mathrm{C} + 273.15 = 303.15\,\mathrm{K} $$
03

Apply the ideal gas law to the initial and final states

Since we are working with constant amounts of the same gas, we can say that \(nR\) is a constant. We will apply the ideal gas law to the initial and final states, and we can then equate the results: $$ P_{i} V_{i} = nR T_{i} $$ $$ P_{f} V_{f} = nR T_{f} $$
04

Eliminate the common term and solve for the final temperature

Notice that both equations have \(nR\) as a common term. We can eliminate \(nR\) by dividing the final state equation by the initial state equation: $$ \frac{P_{f} V_{f}}{P_{i} V_{i}} = \frac{T_{f}}{T_{i}} $$ Now, we can plug in the given values for the volumes and pressures: $$ \frac{(20.0 P_{i})(0.111 V_{i})}{P_{i} V_{i}} = \frac{T_{f}}{303.15 \mathrm{K}} $$ Simplify the expression: $$ \frac{20.0 \cdot 0.111}{1} = \frac{T_{f}}{303.15\, \mathrm{K}} $$ Next, solve for the final temperature \(T_f\): $$ T_{f} = 303.15\, \mathrm{K} \times 20.0 \times 0.111 = 672.27\, \mathrm{K} $$
05

State the final answer

The new temperature of the air after compression is approximately \(672.27\, \mathrm{K}\).

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