Question

A 2-kg mass of helium is maintained at 300 kPa and 278C in a rigid container. How large is the container, in m3?

Answer 1

Answer:

$3.849 m^{3}$

Explanation:

Assuming that helium behaves as an ideal gas and taking gas constant of helium as  

R=2.0769 kJ/KgK

From ideal gas equation

PV=mRT and making V the subject then

$V=\frac {mRT}{P}$

Where m is the mass, V is the volume, P is pressure, T is temperature in Kelvin

Substituting the figures given in the question then

$V=\frac {2\times 2.0769\times 278}{300}=3.849188 \approx 3.849 m^{3}$

Therefore, volume of container is $3.849 m^{3}$