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Posted by on Saturday, April 10, 2010 at 12:16am.

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

  • Physics - , Saturday, April 10, 2010 at 11:17am

    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.

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