Question

A gas contains 0.1 mol of oxygen and .4 moles of Nitrogen. If the sample is at standard temperature and pressure, what is the partial pressure of nitrogen?

Answer 1

Answer: The partial pressure of nitrogen is 1.25 atm

Step-by-step explanation:

From ideal gas equation:

`PV=nRT`

where,

R = Gas constant =
`0.0821\text{ L atm }mol^(-1)K^(-1)`

T = temperature of the gas = 273K (at STP)

P = pressure of the gas = 1 atm (at STP)

n = number of moles = 0.1 +0.4 = 0.5

To calculate the partial pressure of the solution, we use the law given by Dalton, which is:

`P_T=\sum_(i=1)^n (p_i* \chi_i)`

Or,

`P_T=[(p_{\text{N_2}}* \chi_(N_2))`

where,
`\chi_(N_2))`= mole fraction nitrogen =
`\frac{\text {moles of nitrogen}}{\text {total moles}}=(0.4)/(0.5)=0.8`

We are given:

Total pressure = 1 atm

Putting values in above equation, we get:

`1=0.8* p_(N_2)`

`p_(N_2)=1.25atm`

Thus the partial pressure of nitrogen is 1.25 atm