Question

Suppose that Daniel has a 2.50 L bottle that contains a mixture of O, N,, and CO, under a total pressure of 5.60 atm. He knows that the mixture contains 0.310 mol N, and that the partial pressure of Co, is 0.350 atm. If the temperature is 273 K. what is the partial pressure of O,? Pos

Answer 1

Answer: 2.47 atm

Solution: We will use the Dalton's law of partial pressures to solve this question, which states that the total pressure of a mixture of non-reacting gases is equal to the individual partial pressures of all the gases, i.e.

P_total = p_O2 + p_N2 + p_CO2 [equation-1]

Now, partial pressure means the pressure a gas would apply if all of the volume of the gas mixture (i.e. the volume of the bottle) was occupied by that gas only. Therefore, partial pressure of a gas can be calculated from the ideal gas equation which is PV = n_iRT

P = pressure of the gas (here partial pressure)

V = volume of the gas (here, volume of the bottle) = 2.50 L (given)

n_i = number of moles of the i^th gas

R = Universal gas constant = 0.082 L atm K^-1 mol^-1 (from literature) [This particular value of R is chosen as the units will cancel out in the following calculations (see below)]

T = temperature = 273 K (given)

For N₂

p_N2 = (n_N2* RT)/ V

= (0.310 mol * 0.082 L atm K^-1 mol^-1 * 273 K)/ 2.50 L

= 6.94 L atm / 2.50 L

= 2.78 atm

p_CO2 = 0.350 atm (given)

P_total = 5.60 atm (given)

Since, we now have both the partial pressures of N₂ and CO₂ ,and the total pressure is already given, we can easily calculate the partial pressure of O₂ as follows form equation-1:

Partial pressure of O₂, p_O2 = P_total - (p_N2 + p_CO2)

= 5.60 atm - (2.78 atm + 0.350 atm)

= 2.47 atm