Question

A rigid container whose volume can be varied with the use of a piston contains 2.53 mol H₂ gas at STP.

(a) What is the volume of the gas?
_____L

(b) The container is cooled to 209 K. The piston is held fixed so the volume does not change. What is the new pressure inside the container?
____atm

(c) The temperature of the container is held at 209 K and 1.55 mol He gas is added to the container. The pressure inside the container is held constant, but the piston is released so the volume of the container can vary. What is the final volume of the container?
_____ L

(d) What are the partial pressures of H₂ and He in the container in part (c)?

H2: ____
He: ____

(e) The piston on the container in part (c) is set to maintain a constant pressure of 1 atm. The container is then heated from 209 K to 380 K.What is the minimum amount of heat that must be added to the container for it to reach 380 K?
______ J

Answer 1

a) at STP 1 mole gas occupy volume 22.414 liter

then 2.53 mole H₂ occupy volume = 22.414 2.53 = 56.70742 L

b )

use Ideal gas equation

PV = nRT             where, P = atm pressure= ?

V = volume in Liter = 56.70742 L

n = number of mole = 2.53 mol

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 209 K

We can write ideal gas equation

P = nRT/V

Substitute value in above equation

P = 2.530.08205209 / 56.70742 = 0.7650 atm

c)

We know that PV = nRT

V = nRT/P

n = 2.53+1.55 =4.08 mole, T = 209K, P= 0.7650 atm, R = 0.0821 L atm mol^-1 K^-1 ( R = gas constant)

Substitute these value in above equation.

V = 4.080.08205209/0.7650 = 91.4584 L

Volume of gas is 91.4584 L

d) total mole = 4.08 mole

mole fraction of H₂ = 2.53/4.08 = 0.62

mole fraction og He = 1.55/4.08 = 0.38

partial pressure = mole fraction total pressure

partial pressure of H₂ = 0.62 0.7650 = 0.4743 atm

partial pressure of He = 0.38 0.7650 = 0.2907 atm