1) A 9.395 mol sample of xenon gas is maintained in a 0.7940 L container at 300.

Question

1) A 9.395 mol sample of xenon gas is maintained in a 0.7940 L container at 300.8 K. What is the pressure in atm calculated using the van der Waals' equation for Xe gas under these conditions? For Xe, a = 4.194 L2atm/mol2 and b = 5.105×10-2 L/mol.

atm

2) According to the ideal gas law, a 1.010 mol sample of xenon gas in a 1.253 L container at 274.3 K should exert a pressure of 18.14 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure? For Xe gas, a = 4.194 L2atm/mol2 and b =5.105×10-2 L/mol.

%

Hint: % difference = 100×(P ideal - Pvan der Waals) / P ideal

3)According to the ideal gas law, a 9.532 mol sample of argon gas in a 0.8096 L container at 503.0 K should exert a pressure of 486.0 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure? For Ar gas, a = 1.345 L2atm/mol2 and b =3.219×10-2 L/mol.

%

Hint: % difference = 100 × (P ideal - Pvan der Waals) / P ideal

Explanation / Answer

appply van der Waal's equation- ( P + n2 a /V2) ( V - nb ) = nR T

or, P = [ nRT / ( V- nb ) ] - [ n2 a / V2 ]

Substituting n = 9.395 , V = 0.7950 L , T = 300.8 K , a = 4.194 L2 / atm mol2  , b = 5.105 x 10-2 L mol-1  

R = 0.0821L.atm mol-1 K

P = 737.96 - 587.23 = 150.73 atm.

------------------ ---------------------------------------------------------------------------------------------------------------------------------------------

(2) Siimilarly , use van der Wall's equation for this question also, substituting given values as-

n = 1.010,V =1.253 L ,T = 274.3 K ,a = 4.194 L2 atm /mol2 , b = 5.105 x 10-2 L /mol , R = 0.0821 L atm mol1 K to get calculated P (van der Waal's ) = 16.14 atm

Given P (ideal ) = 18.14 atm

So %Differennce = 100 x( P ideal - P van der waal's ) / P ideal

-------------------------- =100 x (18.14 - 16.14) / 18.14 = 11.01%

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

(3) Similarly the third question can also be solved.

Glad to help , but please post Q,(3 ) separately as fresh question for details

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.