Question

At high pressures, real gases do not behave ideally. calculate the pressure exerted by 25.0 g h2 at 20.0°c in a 1.00 l container assuming in part 1 non-ideal behavior and in part 2 ideal behavior.

Answer 1

1. 412atm

2. 300atm

Step-by-step explanation:

Hello,

1. In this case, Van der Waals equation is suitable to model the non-ideal behavior of hydrogen at the specified conditions:

`P=(RT)/(v-b) -(a)/(v^2)`

At the specified conditions, we compute
`a`,
`b` and
`v` as shown below:

`a=(27)/(64)(R^2Tc^2)/(Pc) =(27)/(64)((0.082 (atm*L)/(mol*K))^2*(33.3K)^2)/(12.83atm)=0.245atm(L^2)/(mol^2)\\\\b=(1)/(8)(RTc)/(Pc) =(1)/(8)((0.082 (atm*L)/(mol*K))*(33.3K))/(12.83atm)=0.0266(L)/(mol)\\ \\v=(1.00L)/(25.0gH_2*(1molH_2)/(2gH_2) ) =0.08(L)/(mol)`

Thus, the pressure turns out into:

`P=(0.082 (atm*L)/(mol*K) *293.15K)/(0.08(L)/(mol)-0.0266(L)/(mol))-(0.245atm(L^2)/(mol^2))/((0.08(L)/(mol) )^2) =412atm`

2. In this section, the ideal gas equation is directly applied as shown below:

`P=(nRT)/(V)=((25gH_2*(1molH_2)/(2gH_2) )(0.082 (atm*L)/(mol*K) )(293.15K))/(1.00L)  \\P=300atm`

Best regards.

Answer 2

Van der Waals equation is as follows:
(P + a(n/V)²) * (V -nb) = n RT
Moles of H₂ is calculated by dividing 25 g by 2 (molecular weight of H₂) = 12.5 moles
Values of a and b are:
a = 0.02444 atm L² / mol
b = 0.0266 L / mol
(P + 0.02444 (12.5/1)²) * (1 -(12.5 *0.0266)) = 12.5 * 0.0821 * 293 K
If we solve this equation we get pressure of 412.29 atm

With ideal gas equation we get:
P V = n R T
P = n * R * T / V = (12.5 * 0.0821 * 293 / 1) = 300.69 atm