Determine the temperature, in K, of 5 kg of air at a pressure of 0.3 MPa and a volume of 2.2 m3. Ideal gas behavior can be assumed for air under these conditions.

The temperature is 460.45 K

Explanation:

From the ideal gas equation,

PV = nRT

P is the pressure of air = 0.3 MPa = 0.3*10^6 = 3*10^5 Pa

V is the volume of air = 2.2 m^3

n is the number of moles of air = mass of air/MW of air = 5/29 = 0.1724 kg mol

R is gas constant = 8314.34 m^3. Pa/kg mol. K

T is the temperature of air in Kelvin

T = PV/nR = (3*10^5 * 2.2) / (0.1724 * 8314.34) = 460.45 K