posted by John on .
The boiling point of the helium at i atm is 4.2 k, what is the volume occupied by the helium gas due to evaporation of 30 g of liquid at 1 atm pressure and a temperature of 5.2 k. Answer in L
The boiling point does not affect the calculation. Treat He as an ideal gas at 5.2 K.
30 g is 7.5 moles. That would occupy 7.5*22.4 = 168 liters at 273K and 1 atm. At 5.2K and 1 atm, the volume is 5.2/273 times less, due to the lower temperature.