Question

25.5 moles of helium gas is at 10 0C and gauge pressure of 0.350 ATM. Calculate

a) the volume of the helium (He) gas under these conditions;

b) the temperature of the gas when it is compressed to precisely half of the original volume at the gauge pressure of 1.00 ATM;

c) the mass of the helium gas.

Answer 1

Solution:

a) Given the number of moles = 25.5 moles

=> n = 25.5

Temperature = 283 K

Total pressure = 1 + 0.350 = 1.350 atm

Now Using the ideal gas equation, PV = nRT

=> V = nRT / P

=> V = 25.5 * 8.315 * 283 / 1.350 * 1.013 * 10^5 ( 1 atm = 1.013 * 10^5 Pa)

=> V = 0.439 m^3 (approx)

b) From the question V2 = 0.5 V1 and P2 = 2 atm

=> P1 * V1 / T1 = P2 * V2 / T2

=> T2 = P2 * V2 * T1 / P1 * V1

=> T2 = 2 * 0.5 * V1 * 283 / 1.350 * V1

=> T2 = 1 * 283 / 1.350

=> T2 = 209.63 K

and in Celcius , it will be 209.63 - 273 = -63.37 Celcius

c) malar mass of helium = 4 amu

=> Mass = Molar mass * moles

=> Mass = 4 * 25.5

=> Mass = 102 amu

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