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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 monatomic ideal gas expands isothermally from \(\left\\{p_{1}, V_{1}, T_{1}\right\\}\) to \(\left\\{p_{2}, V_{2}, T_{1}\right\\} .\) Then it undergoes an isochoric process, which takes it from \(\left\\{p_{2}, V_{2}, T_{1}\right\\}\) to \(\left\\{p_{1}, V_{2}, T_{2}\right\\}\) Finally the gas undergoes an isobaric compression, which takes it back to \(\left\\{p_{1}, V_{1}, T_{1}\right\\}\) a) Use the First Law of Thermodynamics to find \(Q\) for each of these processes. b) Write an expression for total \(Q\) in terms of \(p_{1}, p_{2}, V_{1},\) and \(V_{2}\).

Suppose \(15.0 \mathrm{~L}\) of an ideal monatomic gas at a pressure of \(1.50 \cdot 10^{5} \mathrm{kPa}\) is expanded adiabatically (no heat transfer) until the volume is doubled. a) What is the pressure of the gas at the new volume? b) If the initial temperature of the gas was \(300 . \mathrm{K},\) what is its final temperature after the expansion?

One mole of an ideal gas, at a temperature of \(0^{\circ} \mathrm{C}\), is confined to a volume of \(1.0 \mathrm{~L}\). The pressure of this gas is a) \(1.0 \mathrm{~atm}\). b) 22.4 atm. c) \(1 / 22.4 \mathrm{~atm}\) d) \(11.2 \mathrm{~atm}\).

A sample of gas at \(p=1000 . \mathrm{Pa}, V=1.00 \mathrm{~L},\) and \(T=300 . \mathrm{K}\) is confined in a cylinder. a) Find the new pressure if the volume is reduced to half of the original volume at the same temperature. b) If the temperature is raised to \(400 . \mathrm{K}\) in the process of part (a), what is the new pressure? c) If the gas is then heated to \(600 . \mathrm{K}\) from the initial value and the pressure of the gas becomes \(3000 . \mathrm{Pa},\) what is the new volume?

Air at 1.00 atm is inside a cylinder \(20.0 \mathrm{~cm}\) in radius and \(20.0 \mathrm{~cm}\) in length that sits on a table. The top of the cylinder is sealed with a movable piston. A \(20.0-\mathrm{kg}\) block is dropped onto the piston. From what height above the piston must the block be dropped to compress the piston by \(1.00 \mathrm{~mm} ? 2.00 \mathrm{~mm} ? 1.00 \mathrm{~cm} ?\)
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