Question

15.1 L N2 at 25 °C and 125 kPa and 35.9 L O2 at 25 °C and 125 kPa were transferred to a tank with a volume of 5.00 L. What is the total pressure at 41 °C?

Answer 1

Given,

The volume of N₂ gas = 15.1 L

The pressure of N₂ gas = 125 kPa x ( 1 atm / 101.325 kPa) = 1.234 atm

Temperature(T) of N₂ gas = 25 ^oC + 273.15 = 298.15 K

The volume of O₂ gas = 35.9 L

The pressure of O₂ gas = 125 kPa x ( 1 atm / 101.325 kPa) = 1.234 atm

Temperature(T) of O₂ gas  = 25 ^oC + 273.15 = 298.15 K

Volume of tank = 5.00 L

To calculate, the total pressure of gas at 41 ^oC.

Firstly calculating the number of moles of N₂ and O₂ gas,

We know, the ideal gas equation,

PV = nRT

Rearranging the formula,

n = PV /RT ------Here, R = 0.08206 L.atm/mol.K

For N₂,

= (1.234 atm x 15.1 L)/ ( 0.08206 L.atm/mol.K x 298.15K)

= 0.7614 mol N₂

Similarly, For O₂,

= (1.234 atm x 35.9 L)/ ( 0.08206 L.atm/mol.K x 298.15K)

= 1.8102 mol O₂

Now, the total number of moles = Moles of N₂ + Moles of O₂

total number of moles = 0.7614 mol + 1.8102 mol

total number of moles = 2.5716 mol

Now, calculating the total pressure at 41 ^oC,

T = 41 ^oC + 273.15 = 314.15 K

Now, P = nRT /V

P = ( 2.5716 mol x 0.08206 L.atm/mol.K x 314.15 K) / 5.00 L

P = 13.2585 atm

Converting from atm to kPa,

= 13.2585 atm x (101.325 kPa / 1 atm)

= 1343 kPa