The Mendeleev-Clapeyron equation is given by:
PV = nRT
where:
P = pressure (Pa)
V = volume (m^3)
n = number of moles
R = ideal gas constant = 8.314 J/(mol·K)
T = temperature (K)
Given:
n = 2 mol
P = 202 650 Pa
T = 0°C = 273 K (since 0°C = 273 K)
Plugging in the values into the equation:
PV = nRT
V = (nRT) / P
V = (2 mol)(8.314 J/(mol·K))(273 K) / 202650 Pa
V = 1.123 m^3
Therefore, the volume of 2 mol of gas at 0°C and 2 atm is 1.123 m^3.
I. Calculate volume of 2 mol of gas at 0 C and 2 atm = 202 650 Pa using
Mendeleev-Clapeyron equation.
1 answer