Question

If 18.74 mol of helium gas is at 11.5 ?C and a gauge pressure of 0.376 atm .

a) Calculate the volume of the helium gas under these conditions.

b) Calculate the temperature if the gas is compressed to precisely half the volume at a gauge pressure of 1.17 atm .

Answer 1

(a) Use the universal gas law -

PV = nRT rearranging,

=> V = nRT/P

given that -

T = 11.5 degC = 284.5 degK
n = 18.74 moles
R = 8.205 x 10^-5 M^3 - atm/degK-mole
P = 0.376 atm gauge = 1.376 atm absolute

put these values in the above expression -  

V = (18.74 x 8.205 x 10^-5 x 284.5) / 1.376 =  
V = 0.318 M^3

(b) V2 = V1 /2 = 0.318 / 2 = 0.159 M^3
P2 = 2.17 atm.

PV = nRT so T = PV/nR

T = (2.17 x 0.159) / (18.74 x 8.205 x 10^-5) = 224.4 degK = 224.4 - 273 = -48.6 deg C.