Question

The volume of 1.20 lol of gas at 61.3 kPa and 25.0 degree Celsius is

Answer 1

Answer:

V = 48.5 L

Explanation:

Converting °C to K and kPa to atm

T = 25.0°C + 273.15 = 298.15 K

P = 61.3 kPa × (1 atm / 101.325 kPa) = 0.60498 atm

Calculating the volume of gas

V = nRT / P

V = (1.20 mol)(0.082057 L•atm/mol•K)(298.15 K) / 0.60498 atm

V = 48.5 L