Question

An apparatus consists of a 4.0 dm^3 flask containing nitrogen gas at 25°c and 803 kPa, it is joined by a valve to a 10.0dm^3 flask containing argon gas 25°c and 47.2KPa. The valve is opened and the gases mix , what is the partial pressure of each gas and calculate the total pressure of the gas mixture

Answer 1

so total pressure is = 261.92 kPa

partial pressure of Ar gas= 33.50 Kpa

partial pressure of N2 gas = 227 kPa

Step-by-step explanation:

PV = nRT

n = PV / RT

as we know that 1dm3 = 1 Liter

smaller flask before mixing

ideal gas constant

n = (803 kPa) x (4L) / ((8.3144621 L kPa/K mol) x (25 + 273 K))

n = (3212 kPa L) / ((8.3144621 L kPa/K mol) x (298 K))

n = (3212 kPa L) /2477.7

n= 1.29 mol

In the larger flask:

n = (47.2 kPa) x (10 L) / ((8.3144621 L kPa/K mol) x (25 + 273 K)) =

n = (472 kpaL/ 2477.7

n= 0.19 mol

PV = nRT

P = nRT / V

After mixing:

P = (1.29 mol + 0.19 mol) x (8.3144621 L kPa/K mol) x (25+ 273 K) / (4 L + 10 L)

= (1.48) x (2477.7 / (14 L)

= 261.92 kPa total pressure

so total pressure is = 261.92 kPa

a)

(261 kPa) x (0.19 mol Ar) / (1.29 + 0.19mol) =

= 49.59/1.48 = 33.50 Kpa for Arg

b)

(261 kPa total) - (33.50 kPa Ar) = 227 kPa for N2

partial pressure of Ar gas= 33.50 Kpa

partial pressure of N2 gas = 227 kPa