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

Answer:

1. 412atm

2. 300atm

Explanation:

Hello,

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

$P=\frac{RT}{v-b} -\frac{a}{v^2}$

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

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

Thus, the pressure turns out into:

$P=\frac{0.082 \frac{atm*L}{mol*K} *293.15K}{0.08\frac{L}{mol}-0.0266\frac{L}{mol}}-\frac{0.245atm\frac{L^2}{mol^2}}{(0.08\frac{L}{mol} )^2} =412atm$

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

$P=\frac{nRT}{V}=\frac{(25gH_2*\frac{1molH_2}{2gH_2} )(0.082 \frac{atm*L}{mol*K} )(293.15K)}{1.00L} \\P=300atm$

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