The pressure exerted on a sample of gas is 735 mm Hg with a volume of 375 mL and a temperature of 25 degrees Celsius. What would the pressure of the gas be if the temperature was reduced to standard temperature and the volume was expanded to 500.0 mL?

= 498.13 mmHg

Explanation:

Using the combined gas law;

P1V1/T1 = P2V2/T2

in this case; P1 = 725 mmHg, V1 = 375 mL and T1 = 25°C + 273 = 298 K

P2 = ? V2 = 500 mL and T2 (standard temperature) = 0°C + 273 = 273 K

P2 = P1V1T2/T1V2

= (725 * 375 * 273) / (298 * 500)

= 498.13 mmHg