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Maurinko [17]
3 years ago
7

Hydrogen sulfide (H2S) is a gas that smells like rotten eggs. It can be produced by bacteria in your mouth and contributes to ba

d breath. It also causes silver to tarnish so you can actually blacken silver with your breath! Let's use H2S to explore the difference in pressure exerted by an ideal versus a van der Waals gas. Calculate the pressure exerted by 1.00 mol of H2S behaving as
Chemistry
1 answer:
Doss [256]3 years ago
5 0

Answer:

See explanation.

Explanation:

Hello!

In this case, we consider the questions:

a. Ideal gas at:

 i. 273.15 K and 22.414 L.

 ii. 500 K and 100 cm³.

b. Van der Waals gas at:

 i. 273.15 K and 22.414 L.

 ii. 500 K and 100 cm³.

Thus, we define the ideal gas equation and the van der Waals one as shown below:

P^{id}=\frac{nRT}{V}\\\\ P^{vdW}=\frac{RT}{v-b}-\frac{a}{v^2}

Whereas b and a for hydrogen sulfide are 0.0434 L/mol and 4.484 L²*atm / mol² respectively, therefore, we proceed as follows:

a.

 i. 273.15 K and 22.414 L.

P^{id}=\frac{1.00mol*0.08206\frac{atm*L}{mol*K}*273.15K}{22.414L}=1 .00 atm

 ii. 500 K and 100 cm³ (0.1 L).

P^{id}=\frac{1.00mol*0.08206\frac{atm*L}{mol*K}*500K}{0.100L}=410.3 atm

b.

 i. 273.15 K and 22.414 L: in this case, v = 22.414 L / 1.00 mol = 22.414 L/mol

P^{vdW}=\frac{0.08206\frac{atm*L}{mol*K}*273.15K}{22.414L/mol-0.0434L/mol}-\frac{4.484 atm*L^2/mol^2}{(22.414L/mol)^2}=0.993atm

 ii. 500 K and 100 cm³: in this case, v = 0.1 L / 1.00 mol = 0.100 L/mol

P^{vdW}=\frac{0.08206\frac{atm*L}{mol*K}*500K}{0.100L/mol-0.0434L/mol}-\frac{4.484 atm*L^2/mol^2}{(0.100L/mol)^2}=276.5atm

Whereas we can see a significant difference when the gas is at 500 K and occupy a volume of 0.100 L.

Best regards!

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