A sample consisting of 1.00 mol of the molecules in air is expanded isothermally
ID: 894410 • Letter: A
Question
A sample consisting of 1.00 mol of the molecules in air is expanded isothermally at 25 degree C from 24.2 dm3 to 48.4 dm^3 (a) reversibly, (b) against a constant external pressure equal to the final pressure of the gas, and (c) freely [against zero external pressure). For the three processes calculate q, w, Delta U, and Delta H. In an industrial process, nitrogen is heated to 500 K at a constant volume of 1.000 m3. The gas enters the container at 300 K and 100 atm. The mass of the gas is 92.4 kg. Use the van der Waals equation to determine the approximate pressure of the gas at its working temperature of 500 K. For nitrogen. a = 1.352 dm^6 atm mol^2 h = 0.03R7 dm^3 mol^-1.Explanation / Answer
for isothermal reversible expansion
DelU ( Change in internal energy ) = delH ( Change in enthalpy) = 0
Q= RTln(V2/V1)
where V2= 48.2 dm3
V1= 24.2 dm3
T= 25 Deg.C =25+273.15= 298.15 K
Q= 8.314* 298.15* ln (48.4/24.2)= 1718.186 J/mol.K
for one mole Q = 1718.186*1 =1718.186 Joules
Q= -W =-1718.186 joules
For expansion against constant external pressure under isothermal conditions
DelU= DelH= O
W= -P*delV
P = constant external pressure = nRT/ V= 1* 8.314 *1000 *298/48.4 Pa = 51189.5 Pa
W= -51189.5 * 24.2 dm3/1000 = 1238.76 Joules
from first law Q= DelU- W= 1238.76 Joules
c) during free expansion W= 0, isothermal condition DeU= DelH= O
and Q= Delu-W = 0
b) For vandewaal gas (P + an²/V²) x (V - nb) = nRT
n= 92.4/28= 3.3 kg moles =3.3*1000 gmoles
R= 0.08206 L.atm/mole.K
a= 1.352 dm6atm./ mole2 =1.352 L6atm/mole2
P = [nRT / (V - nb)] - an²/V² = 3300*0.08206*500/(1000-3300*0.0387)- 1.352*(3300)2/ 106= 140 atm
b= 0.0387 dm3/mol =0.0387 L/mole
T=500 K
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