/*! 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 41 A cylinder in a car engine takes... [FREE SOLUTION] | 91影视

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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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Most popular questions from this chapter

A \(10.0-\mathrm{L}\) vessel contains \(12 \mathrm{g}\) of \(\mathrm{N}_{2}\) gas at \(20^{\circ} \mathrm{C}\) (a) Estimate the nearest-neighbor distance. (b) Is the gas dilute? [Hint: Compare the nearest-neighbor distance to the diameter of an \(\left.\mathrm{N}_{2} \text { molecule, about } 0.3 \mathrm{nm} .\right]\)
A square brass plate, \(8.00 \mathrm{cm}\) on a side, has a hole cut into its center of area \(4.90874 \mathrm{cm}^{2}\) (at $\left.20.0^{\circ} \mathrm{C}\right) .$ The hole in the plate is to slide over a cylindrical steel shaft of cross-sectional area \(4.91000 \mathrm{cm}^{2}\) (also at \(20.0^{\circ} \mathrm{C}\) ). To what temperature must the brass plate be heated so that it can just slide over the steel cylinder (which remains at \(20.0^{\circ} \mathrm{C}\) )? [Hint: The steel cylinder is not heated so it does not expand; only the brass plate is heated. \(]\)
The SR- 71 Blackbird reconnaissance aircraft is primarily made of titanium and typically flies at speeds above Mach \(3 .\) In flight, the length of the SR- 71 increases by about \(0.20 \mathrm{m}\) from its takeoff length of $32.70 \mathrm{m} .$ The average coefficient of linear expansion for titanium over the temperature range experienced by the \(\mathrm{SR}-71\) is $10.1 \times 10^{-6} \mathrm{K}^{-1} .$ What is the approximate temperature of the SR-71 while it is in flight if it started at \(20^{\circ} \mathrm{C} ?\)
What is the mass density of air at \(P=1.0\) atm and $T=(a)-10^{\circ} \mathrm{C}\( and \)(\mathrm{b}) 30^{\circ} \mathrm{C} ?$ The average molecular mass of air is approximately 29 u.
A diver rises quickly to the surface from a 5.0 -m depth. If she did not exhale the gas from her lungs before rising, by what factor would her lungs expand? Assume the temperature to be constant and the pressure in the lungs to match the pressure outside the diver's body. The density of seawater is $1.03 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}$
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