Question

Compute the mass of a 2 m³ propane at 280 kpa and 40˚C.

m = ____ kg

Answer 1

We have to use the ideal gas law (PV=nRT, where P is the pressure, V the volume (V=2 m³),

T is the thermodynamic temperature (T = 40 °C = 313.15 K) and n is the mole number

equal to the quotient between the mass of propane and the molar mass (M=44.09 g/mol).

After using the proper conversions and R = 8.3144 Pa m³/molK = 0.0083144 kPa m³/molK,

we find the amount of propane gas as it follows:

$PV=nRT \iff n=\cfrac{PV}{RT}=\cfrac{m}{M} \\ \implies m=\cfrac{PVM}{RT} \\ \text{ } \\m= \cfrac{(280\,k{Pa})(2\,m^3)(0.04409\,kg/{mol})}{(0.0083144\frac{kPa\cdot m^3}{mol \cdot K})(313.15\,K)} \\ \text{ } \\m=9.48\,{kg}$

In conclusion, the mass of propane gas is 9.48 kg.

Reference

• Chang, R., & Goldsby, K. A. (2010). Chemistry. Chemistry, 10th ed.; McGraw-Hill Education: New York, NY, USA.