Question

A sample of gas for which \(p=1000 . \mathrm{Pa}, V=1.00 \mathrm{~L}\), 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 \(600 .\) K from the initial value and the pressure of the gas becomes \(3000 . \mathrm{Pa}\), what is the new volume?

Short answer

Question: A gas initially occupies a volume of 1.00 L at a pressure of 1000 Pa and a temperature of 300 K. Calculate the following: (a) The new pressure when the volume is reduced to half its original volume while maintaining a constant temperature. (b) The new pressure when the volume is reduced to half its original volume and the temperature is increased to 400 K. (c) The new volume, given that the pressure is increased to 3000 Pa and the temperature is increased to 600 K. Answer: (a) The new pressure after the volume is reduced to half its original volume at constant temperature is 2000 Pa. (b) The new pressure after the volume is reduced to half its original volume and the temperature is increased to 400 K is approximately 2666.67 Pa. (c) The new volume when the pressure is increased to 3000 Pa and the temperature is increased to 600 K is approximately 0.66667 L.

Step-by-step solution

Part (a) - New pressure for half the original volume at constant temperature

Use Boyle's Law (\(p_1 V_1 = p_2 V_2\)) since the temperature is constant. The original pressure (\(p_1\)) is 1000 Pa, the original volume (\(V_1\)) is 1.00 L and the new volume (\(V_2\)) is half of the original volume, which is 0.50 L. We need to find the new pressure (\(p_2\)). Solve for \(p_2\): \(p_2 = \frac{p_1 V_1}{V_2} = \frac{1000 \ \text{Pa} \times 1.00 \ \text{L}}{0.50 \ \text{L}} = 2000\ \text{Pa}\) The new pressure is \(2000\, \text{Pa}\).

Part (b) - New pressure for half the original volume with increased temperature

Use the Combined Gas Law \({\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}}\). The initial state has a pressure of 1000 Pa (\(p_1\)), a volume of 1.00 L (\(V_1\)) and a temperature of 300 K (\(T_1\)). The new state has a volume of 0.50 L (\(V_2\)) and a temperature of 400 K (\(T_2\)). We need to find the new pressure (\(p_2\)). Solve for \(p_2\): \(p_2 = \frac{p_1 V_1 \times T_2}{T_1 \times V_2} = \frac{1000 \ \text{Pa} \times 1.00 \ \text{L} \times 400\ \text{K}}{300\ \text{K} \times 0.50 \ \text{L}} = 2666.67\ \text{Pa}\) The new pressure is approximately \(2666.67\, \text{Pa}\).

Part (c) - New volume for increased temperature and pressure

Use the Combined Gas Law \({\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}}\). The initial state has a pressure of 1000 Pa (\(p_1\)), a volume of 1.00 L (\(V_1\)) and a temperature of 300 K (\(T_1\)). The new state has a pressure of 3000 Pa (\(p_2\)) and a temperature of 600 K (\(T_2\)). We need to find the new volume (\(V_2\)). Solve for \(V_2\): \(V_2 = \frac{p_1 V_1 \times T_2}{T_1 \times p_2} = \frac{1000 \ \text{Pa} \times 1.00 \ \text{L} \times 600\ \text{K}}{300\ \text{K} \times 3000\ \text{Pa}} = 0.66667\ \text{L}\) The new volume is approximately \(0.66667\, \text{L}\).