Question

An unknown volume of laughing gas has a temperature of 325 K and a pressure of 0.5 kPa. The temperature is decreased to 320 K, the pressure is increased to 1.2 kPa, and the final volume of the gas is 48.0 L. What was the initial volume of the laughing gas?

Answer 1

Since pressure, volume, and temperature are changing, use the combined gas law:\[\frac{P_1*V_1}{T_1} = \frac{P_2*V_2}{T_2}\]rearrange and solve for V1