Question

A 7.95 L

container holds a mixture of two gases at 25 °C.
The partial pressures of gas A and gas B, respectively, are 0.352 atm
and 0.715 atm.
If 0.240 mol
of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

To solve this problem, we can use the ideal gas law equation:

PV = nRT

where:
P = pressure
V = volume = 7.95 L
n = number of moles
R = gas constant = 0.08206 L atm K^-1 mol^-1
T = temperature = 25°C = 298 K

We can start by calculating the total number of moles of gas in the container before the third gas is added:

n_total = (P_A + P_B) V/RT

n_total = [(0.352 atm) + (0.715 atm)](7.95 L)/(0.08206 L atm K^-1 mol^-1)(298 K)

n_total = 6.06 mol

We know that 0.240 mol of a third gas is added, so the total number of moles of gas becomes:

n_total = 6.06 mol + 0.240 mol

n_total = 6.30 mol

Now we can use the ideal gas law to calculate the total pressure of the mixture:

P_total = n_total RT/V

P_total = (6.30 mol)(0.08206 L atm K^-1 mol^-1)(298 K)/(7.95 L)

P_total = 0.239 atm

Therefore, the total pressure of the mixture after the third gas is added is 0.239 atm.

:)’