Question

A 7.0-liter sample of N2(g) at 3.50 atm, a 2.5-liter sample of Ne(g) at 1.00 atm, and a 5.0-liter sample of H2(g) at 2.45 atm are transferred to a container. The temperature remains constant throughout at 25 °C. The concentration of the mixture in the container is found to be 0.160 mol/L. Calculate the partial pressure of each gas and the total final pressure in the container.

Answer 1

Answer:

Total pressure = 6.95 atm

pN2)  4.34 atm

p(Ne) =0.44 atm

p(H2) =  2.18 atm

Explanation:

Step 1: Data given

The N2 sample = 7.0 liter at 3.50 atm

The Ne sample = 2.5 liter at 1.00 atm

The H2 sample = 5.0 L at 2.45 atm

Temperature = 25.0 °C

Concentration = 0.160 mol/L

Step 2: Calculate moles

p*V = n*R*T

n = (p*V)/(R*T)

⇒ n = the number of moles of gas

⇒ V = the volume of gas

⇒ p = the pressure of gas

⇒ R = the gas constant

⇒T = the temperature

n(N2) = (3.50*7.0)/(0.08206*298 K)

n(N2) = 1.00 moles

n(Ne) = (1.00*2.5)/(0.08206*298)

n(Ne) = 0.102 moles

n(H2) = (2.45 * 5.0) / (0.08206*298)

n(H2) = 0.501 moles

Step 3: Calculate mol fraction

Total moles = 1.00 + 0.102 + 0.501 = 1.603 moles

Step 4: Calculate mol fraction

Mol fraction = moles gas / total moles

Mol fraction N2 = 1.00 / 1.603 = 0.624

Mol fraction Ne = 0.102/1.603 = 0.0636

Mol fraction H2 = 0.501 / 1.603 = 0.313

Step 5: Calculate total pressure

Total pressure = 3.50 atm + 1.00 atm + 2.45 atm

Total pressure = 6.95 atm

Step 6: Calculate partial pressure

pN2) = 0.624 * 6.95 = 4.34 atm

p(Ne) = 0.0636 *6.95 =0.44 atm

p(H2) = 0.313 *6.95 = 2.18 atm