Question

A closed container contains a volume V of gas, composed of a mixture of 2 moles of O2 and 3 moles of CO2. The total pressure inside the container is 900 mm Hg. The container is at a temperature of 37⁰C? 1. Calculate PO2 and PCO2, the partial pressures of O2 and CO2 in the gas mixture. 2. Calculate the density of the gas mixture. 3. A quantity of water is introduced into the container, keeping the same quantity of gas, as before, above the water. The container is fitted with a piston, which is positioned so that the gaseous phase occupies a volume V′ and the total pressure of this gaseous phase above the water is maintained at 900 mm Hg. what is the difference between the gaseous phases in the two situations?

Answer 1

We have a closed container, so the number of moles remains constant.

Now, the total pressure is equal to the sum of the partial pressures. And the partial pressure of a gas will depend on its molar fraction, that is, the moles of the gas over the total moles. So the partial pressure is defined as:

`P_i=(n_i)/(n_T)P_T`

Where,

Pi is the partial pressure of the gas

ni, are the moles of the corresponding gas

nT, are the total moles

Pt is the total pressure.

1. Partial pressures of O2 and CO2

`\begin{gathered} P_(O2)=(2molO_2)/(2molO_2+3molCO_2)*900mmHg \\ P_(O2)=360mmHg \end{gathered}`
`\begin{gathered} P_(CO2)=(3molCO_2)/(2molO_2+3molCO_2)*900mmHg \\ P_(CO2)=540mmHg \end{gathered}`

Answer: PO2=360mmHg

PCO2=540mmHg

2. Now, the density will depend of the number of moles and the volume. We can calculate the volume of the gases with the ideal gas equation that says:

`\begin{gathered} PV=nRT \\ V=(nRT)/(P) \end{gathered}`

For each gas we will have:

O2

PO2=340mmHg=0.45atm

T=37°C=310.15K

R is a constant = 0.08206 (atm.L)/(mol.K)

nO2=2mol

massO2=2mol x MolarMass = 2 mol x 31.998g/mol=63.996g

`V_(O2)=(2mol*0.08206(atm.L)/(mol.K)*310.15K)/(0.45atm)=113.1L`
`DensityO_2=(Mass)/(Volume)=(63.996g)/(113.1L)=0.57g/L`

CO2

PCO2=540mmHg=0.71atm

T=37°C=310.15K

R is a constant = 0.08206 (atm.L)/(mol.K)

nCO2=3mol

massCO2=3mol x MolarMass = 3 mol x 44.01g/mol=132.03g

`V_(CO2)=(3mol*0.08206*(atm)/(mol)*310.15K)/(0.71atm)=107.5L`
`DensityO_2=(Mass)/(Volume)=(132.03g)/(107.5L)=1.23g/L`

Answer:

Density O2=0.57g/L

Density CO2 = 1.23g/L

3. In the second situation, what will happen is that the partial pressure of the gases will decrease, since due to the pressure exerted by the piston, part of the moles of gas will dissolve in the water. So in the gas phase we will have fewer moles of gas.