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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 tire on a car is inflated to a gauge pressure of \(32 \mathrm{lb} / \mathrm{in}^{2}\) at a temperature of \(27^{\circ} \mathrm{C}\). After the car is driven for \(30 \mathrm{mi}\), the pressure has increased to \(34 \mathrm{lb} / \mathrm{in}^{2} .\) What is the temperature of the air inside the tire at this point? a) \(40^{\circ} \mathrm{C}\) b) \(23^{\circ} \mathrm{C}\) c) \(32^{\circ} \mathrm{C}\) d) \(54^{\circ} \mathrm{C}\)

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