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 nitrogen (V) = 15.1 L

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

Pressure(P) = 125 kPa x ( 1000 Pa / 1 kPa) x ( 1 atm / 101325 Pa) = 1.234 atm

Also given,

The volume of oxygen (V) = 35.9 L

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

Pressure(P) = 125 kPa x ( 1000 Pa / 1 kPa) x ( 1 atm / 101325 Pa) = 1.234 atm

Calculating the number of moles of N₂ and O₂ from the given data,

We know, the ideal gas equation,

PV = nRT

Rearranging the formula,

n = PV/RT

For N₂ ,

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

n = 0.7614 mol N₂

Similarly, for O₂,

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

n = 1.8102 mol O₂

Now,

total moles = Moles of N₂ + Moles of O₂

total moles = 0.7614 mol + 1.8102 mol

total moles = 2.5716 mol

Now, the new volume = 5.00 L

New temperature(T) = 41 ^oC + 273.15 = 314.15 K

Now, rearranging the ideal gas equation,

P = nRT / V

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

P = 13.26 atm

Now, converting atm to kPa,

= 13.26 atm x ( 101325 Pa / 1 atm) x ( 1 kPa / 1000 Pa)

= 1343.4 kPa

Thus, the total pressure = 1.34 x 10³ kPa