Question

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

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.

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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