Question

What pressure is exerted by a mixture of 2.10 g of H₂ and 8.700 g of N₂ at 273°C in a 10.0 L container?

Answer 1

Answer:-

Given:-

volume of container (V) = 10.0 L

temperature (T) = 273 ⁰C = 273 + 273 = 546 K

wt. of H₂ gas = 2.10 g

wt. of N₂ gas = 8.700 g

partial pressure of H₂ gas (P_H2) = ?

partial pressure of N₂ gas (P_N2) = ?

total pressure exerted by a mixture of H₂ and N₂ gases in the container (P_total) =?

As we know that

molar mass of H₂ gas = 2 $\times$ molar mass of H

molar mass of H₂ gas = 2 $\times$ 1

molar mass of H₂ gas = 2 g /mol

similarly

molar mass of N₂ gas = 2 $\times$ molar mass of N

molar mass of N₂ gas = 2 $\times$ 14

molar mass of N₂ gas = 28 g /mol

So

no. of moles of H₂ gas (n_H2) = wt. of H₂ gas / molar mass of H₂ gas

no. of moles of H₂ gas (n_H2) = 2.10 g / 2 g /mol

no. of moles of H₂ gas (n_H2) = 1.05 mol

similarly

no. of moles of N₂ gas (n_N2) = wt. of N₂ gas / molar mass of N₂ gas

no. of moles of N₂ gas (n_N2) =  8.700 g / 28 g /mol

no. of moles of N₂ gas (n_N2) = 0.311 mol

As we know that

gas constant (R) = 0.08206 L.atm.K^-1mol^-1

So according to formula

PV = nRT

P = nRT / V

therefore

partial pressure of H₂ gas (P_H2) = no. of moles of H₂ gas (n_H2) $\times$ gas constant (R) $\times$ temperature (T) / volume of container (V)

partial pressure of H₂ gas (P_H2) = 1.05 mol $\times$ 0.08206 L.atm.K^-1mol^-1$\times$ 546 K / 10.0 L

partial pressure of H₂ gas (P_H2) = 47.044998‬ L.atm / 10.0 L

partial pressure of H₂ gas (P_H2) = 4.70 atm

similarly

partial pressure of N₂ gas (P_N2) = no. of moles of N₂ gas (n_H2) $\times$ gas constant (R) $\times$ temperature (T) / volume of container (V)

partial pressure of N₂ gas (P_N2) = 0.311 mol $\times$ 0.08206 L.atm.K^-1mol^-1$\times$ 546 K / 10.0 L

partial pressure of N₂ gas (P_N2) = 13.93428036‬ L.atm / 10.0 L

partial pressure of N₂ gas (P_N2) = 1.39 atm

Also we know that according to the Dalton's Law of partial pressure " the total pressure of mixture of gases is equal to the sum of all the partial pressures of gases present in the mixture".

therefore

total pressure exerted by a mixture of H₂ and N₂ gases in the container (P_total) = partial pressure of H₂ gas (P_H2) + partial pressure of N₂ gas (P_N2)

total pressure exerted by a mixture of H₂ and N₂ gases in the container (P_total) = 4.70 atm  + 1.39 atm

total pressure exerted by a mixture of H₂ and N₂ gases in the container (P_total) = 6.09 atm (i.e the answer)