Question

A 9.10 L

container holds a mixture of two gases at 43 °C.
The partial pressures of gas A and gas B, respectively, are 0.264 atm
and 0.548 atm.
If 0.130 mol
of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

we can use the ideal gas law equation:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

First, we can find the total number of moles of gas in the container:

n(total) = n(A) + n(B)

To find the number of moles of each gas, we can use the partial pressures and the total pressure:

P(A) / P(total) = n(A) / n(total)

P(B) / P(total) = n(B) / n(total)

We can rearrange these equations to solve for n(A) and n(B):

n(A) = P(A) / P(total) × n(total)

n(B) = P(B) / P(total) × n(total)

We know that the partial pressures of gas A and gas B are 0.264 atm and 0.548 atm, respectively, and we can find the total pressure by adding these partial pressures:

P(total) = P(A) + P(B) = 0.264 atm + 0.548 atm = 0.812 atm

We can also find the total number of moles of gas in the container:

n(total) = PV / RT = (0.812 atm) × (9.10 L) / (0.08206 L·atm/mol·K × 316 K) = 0.286 mol

Now we can add 0.130 mol of a third gas, which gives us a new total number of moles of gas:

n(new) = n(total) + 0.130 mol = 0.286 mol + 0.130 mol = 0.416 mol

Since there is no change in volume or temperature, the new total pressure will be proportional to the total number of moles of gas:

P(new) = P(total) × n(new) / n(total) = (0.812 atm) × (0.416 mol) / (0.286 mol) = 1.18 atm

Therefore, the total pressure will become 1.18 atm.