1: according to the ideal gas law, a 1.016 mol sample of Krypton gas is in a 1.8

Question

1:
according to the ideal gas law, a 1.016 mol sample of Krypton gas is in a 1.875 liter container at 273.8 Kelvin should exert a pressure of 12.17 atmospheres by what percent does the pressure calculated using the boundary waals equation differ from the ideal pressure?
for Kr gas, a=2.318 L^2 ATM/mol^2 and b=3.978x10^-2

2:

the average molecular speed in a sample of O3 gas at a certain temperature is 399 m/s.
the average molecular speed in a sample of New gas is _____ m/s at the same temperature.

Explanation / Answer

Vanderwaal gas equation is

(P+an2/V2)*(V-nb)= nRT , where if P is atm, V= in Liters, R =0.0821 L.atm/mole.K

P= nRT/(V-nb) - an2/V2, (1)

given V= 1.875 L, T= 273.8 K, n= 1.016 moles, a =2.318 L2atm/mole2, b =3.978*10-2 L/mole

substituting these values in Eq.1

P= 1.016*0.0821*273.8/(1.875-1.016*3.978*10-2)- 2.318*1.016*1.016/(1.875*1.875)=11.77 atm

% of pressure difference =100*{ (ideal gas pressure- pressure using vanderwaal equation)/ideal gas pressure}

=100*(12.1-11.17)/12.17 =7.64%

2. let M1= molar mass of O3 gas =48 and M2= molar mass of New gas

V1= average speed of O3= 399 m/s and V2= average speed of New gas

average speed, V= sqrt(3RT/M)

at constant temperature for two different gases

V2/V1=sqrt(M1/M2)

V2= V1*sqrt(M1/M2)= 399*sqrt( 48/M2) m/s

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