Question

a mixture was made using 5.0 atm of oxygen gas and 1600 torr of nitrogen gas in a 14L container at 20 degrees celsius

what is total pressure in the system?

next, the volume of the mixture above was reduced to 8.3L while the temperature was increased to 230 degrees celsius what is the new pressure of the mixture?

Answer 1

The total pressure of a mixture of gases can be defined as the sum of the pressures of each individual gas: P_total=P₁+P₂+… +P_n.

P_{total =} P_O2 + P_N2

1 atm = 760 torr

so, 1600 torr of N₂= (1/760) x 1600 atm = 2.10 atm

P_total= 5.0 atm + 2.10 atm

P_total= 7.10 atm

2.  Combined Gas Law

(P₁)(V₁)/T₁ = (P₂)(V₂) T₂ ........................(i)

So, Initial volume V₁= 14 L

Initial temperature T_{1 =} 20° C = 293 K

Initial pressure = 7.10 atm

Final volume = 8.3L

Final temperature = 230° C = 503 K

Final pressue = (P₂)

so , from eqn (i)

(7.10 atm x 14 L)/293 K = (P₂ x 8.3 L)/503 K

P₂ = 20.55 atm

Answer 2

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