Question

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

Answer 1

See explanation.

Step-by-step 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)=(nRT)/(V)\\\\ P^(vdW)=(RT)/(v-b)-(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)=(1.00mol*0.08206(atm*L)/(mol*K)*273.15K)/(22.414L)=1 .00 atm`

ii. 500 K and 100 cm³ (0.1 L).

`P^(id)=(1.00mol*0.08206(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)=(0.08206(atm*L)/(mol*K)*273.15K)/(22.414L/mol-0.0434L/mol)-(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)=(0.08206(atm*L)/(mol*K)*500K)/(0.100L/mol-0.0434L/mol)-(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!