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Chapter 19: Problem 70

A gas expands at constant pressure from \(3.00 \mathrm{~L}\) at \(15.0^{\circ} \mathrm{C}\) until the volume is \(4.00 \mathrm{~L}\). What is the final temperature of the gas?

Short Answer

Expert verified
Answer: The final temperature of the gas is \(384.2 \mathrm{~K}\).

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given initial temperature from Celsius to Kelvin using the formula: K = °C + 273.15 Initial temperature = \(15 ^\circ \mathrm{C}\) \(T_1 = 15 + 273.15 = 288.15 \mathrm{~K}\)
02

Set up Charles' Law formula

Using the Charles' Law formula for gas under constant pressure: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) Given: \(V_1 = 3.00 \mathrm{~L}\), \(V_2 = 4.00 \mathrm{~L}\), and \(T_1 = 288.15 \mathrm{~K}\) Our goal is to find the final temperature, \(T_2\).
03

Substitute the values and solve for the final temperature

Now we'll substitute the given values into the formula and solve for the final temperature: \(\frac{3.00 \mathrm{~L}}{288.15 \mathrm{~K}} = \frac{4.00 \mathrm{~L}}{T_2}\) To solve for \(T_2\), we'll cross-multiply and divide: \(T_2 = \frac{4.00 \mathrm{~L} \cdot 288.15 \mathrm{~K}}{3.00 \mathrm{~L}}\)
04

Calculate the final temperature

Now, we'll perform the calculations: \(T_2 = \frac{4.00 \cdot 288.15}{3.00}\) \(T_2 = 384.2 \mathrm{~K}\) The final temperature of the gas is \(384.2 \mathrm{~K}\).

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Most popular questions from this chapter

A diesel engine works at a high compression ratio to compress air until it reaches a temperature high enough to ignite the diesel fuel. Suppose the compression ratio (ratio of volumes) of a specific diesel engine is 20 to \(1 .\) If air enters a cylinder at 1.00 atm and is compressed adiabatically, the compressed air reaches a pressure of 66.0 atm. Assuming that the air enters the engine at room temperature \(\left(25.0^{\circ} \mathrm{C}\right)\) and that the air can be treated as an ideal gas, find the temperature of the compressed air.

One hundred milliliters of liquid nitrogen with a mass of \(80.7 \mathrm{~g}\) is sealed inside a 2 - \(\mathrm{L}\) container. After the liquid nitrogen heats up and turns into a gas, what is the pressure inside the container? a) 0.05 atm b) 0.08 atm c) \(0.09 \mathrm{~atm}\) d) 9.1 atm e) 18 atm

1 .00 mol of molecular nitrogen gas expands in volume very quickly, so no heat is exchanged with the environment during the process. If the volume increases from \(1.00 \mathrm{~L}\) to \(1.50 \mathrm{~L},\) determine the work done on the environment if the gas's temperature dropped from \(22.0^{\circ} \mathrm{C}\) to \(18.0^{\circ} \mathrm{C}\). Assume ideal gas behavior.

Chapter 13 examined the variation of pressure with altitude in the Earth's atmosphere, assuming constant temperature-a model known as the isothermal atmosphere. A better approximation is to treat the pressure variations with altitude as adiabatic. Assume that air can be treated as a diatomic ideal gas with effective molar mass \(M_{\text {air }}=28.97 \mathrm{~g} / \mathrm{mol}\) a) Find the air pressure and temperature of the atmosphere as functions of altitude. Let the pressure at sea level be \(p_{0}=101.0 \mathrm{kPa}\) and the temperature at sea level be \(20.0^{\circ} \mathrm{C}\) b) Determine the altitude at which the air pressure and density are half their sea-level values. What is the temperature at this altitude, in this model? c) Compare these results with the isothermal model of Chapter \(13 .\)

As noted in the text, the speed distribution of molecules in the Earth's atmosphere has a significant impact on its composition. a) What is the average speed of a nitrogen molecule in the atmosphere, at a temperature of \(18.0^{\circ} \mathrm{C}\) and a (partial) pressure of \(78.8 \mathrm{kPa} ?\) b) What is the average speed of a hydrogen molecule at the same temperature and pressure?
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