# If 5.0 of nitrogen gas and 5.0 g of chlorine gas are injected in to a 2.0 L vessel at a temperature of 65"^oC, what will the partial pressure of each gas be? What will the total pressure in the container be?

Sep 30, 2016

${p}_{\text{nitrogen}}$ $\cong$ $2.5 \cdot a t m$; ${p}_{\text{chlorine}}$ $\cong$ $1.0 \cdot a t m$.

${p}_{\text{total}}$ $\cong$ $3.5 \cdot a t m$

#### Explanation:

Dalton's law of partial pressure holds that in a gaseous mixture the pressure exerted by a component gas is the same as if that gas ALONE occupied the container; the total pressure is the sum of the individual partial pressures.

We have the parameters to solve the ideal gas equation individually. We know we must specify an absolute temperature. We also know that both chlorine and dinitrogen are diatomic gases (in fact all the elemental gases, save the Noble Gases, and mercury, are diatomic)

i.e. $P = \frac{n R T}{V}$

${p}_{\text{nitrogen}}$ $=$ $\frac{\left(\frac{5.0 \cdot \cancel{g}}{28.02 \cdot \cancel{g} \cdot \cancel{m o {l}^{-} 1}}\right) \times 0.0821 \cdot \cancel{L} \cdot a t m \cdot \cancel{{K}^{-} 1} \cdot \cancel{m o {l}^{-} 1} \times 338 \cancel{K}}{2.0 \cdot \cancel{L}}$ $=$ $2.48 \cdot a t m$.

${p}_{\text{chlorine}}$ $=$ $\frac{\left(\frac{5.0 \cdot \cancel{g}}{70.9 \cdot \cancel{g} \cdot \cancel{m o {l}^{-} 1}}\right) \times 0.0821 \cdot \cancel{L} \cdot a t m \cdot \cancel{{K}^{-} 1} \cdot \cancel{m o {l}^{-} 1} \times 338 \cancel{K}}{2.0 \cdot \cancel{L}}$ $=$ $0.98 \cdot a t m$.

${P}_{\text{Total}}$ $=$ ${p}_{\text{nitrogen"+p_"chlorine}}$ $=$ ??atm