Question

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

Question: Given the initial volume of a gas of 3.00 L and an initial temperature of 15.0°C, find the final temperature when the gas expands to a volume of 4.00 L at constant pressure. Answer: The final temperature of the gas is 384.20 K.

Step-by-step solution

Convert temperatures to Kelvin

Before we start solving the problem, it's important to convert the given temperatures from Celsius to Kelvin because the Ideal Gas Law uses Kelvin as the unit for temperature. To do this, we add 273.15 to the Celsius temperature. Initial temperature in Celsius: \(T1 = 15.0^{\circ}\mathrm{C}\) Convert to Kelvin: \(T1 = 15.0 + 273.15 = 288.15 \mathrm{K}\)

Apply Charles's Law

Charles's Law states that the volume of an ideal gas at constant pressure is directly proportional to its absolute temperature. Mathematically, it can be represented as: $$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$ where \(V_1\) and \(T_1\) are the initial volume and temperature of the gas, and \(V_2\) and \(T_2\) are the final volume and temperature. We are given \(V_1 = 3.00 \mathrm{~L}\), \(T_1 = 288.15 \mathrm{K}\), and \(V_2 = 4.00 \mathrm{~L}\). We can now solve for \(T_2\) by rearranging the formula: $$T_2 = \frac{V_2 \times T_1}{V_1}$$

Calculate the final temperature

Now, we can plug the values into the formula and solve for \(T_2\): $$T_2 = \frac{4.00 \mathrm{~L} \times 288.15 \mathrm{K}}{3.00 \mathrm{~L}}$$ $$T_2 = 384.20 \mathrm{K}$$ The final temperature of the gas is \(384.20 \mathrm{K}\).