Question

A gas mixture contains oxygen and argon at partial pressures of $0.60atm$and $425mmHg$. If nitrogen gas added to the sample increase the total pressure to $1250torr$, what is the partial pressure, in torr, of the nitrogen added?

Short answer

The partial pressure of nitrogen is$369torr$.

Step-by-step solution

Given Information

We have given the pressure of oxygen $0.60atm$and argon is$425mmHg$.

Calculation

Considering the given conditions.

The partial pressure of oxygen $$\begin{gathered} p\left(O_2\right)=0.60atm=\left(0.60\times 760\right)torr \\ p\left(O_2\right)=456torr \end{gathered}$$,

the partial pressure of argon $p\left(Ar\right)=425torr$and

total partial pressure $p\left(total\right)=1250torr$

Now, calculating the partial pressure of nitrogen

$$\begin{gathered} p\left(N_2\right)=p\left(total\right)-p\left(Ar\right)-p\left(O_2\right) \\ p\left(N_2\right)=1250torr-425torr-456torr \\ p\left(N_2\right)=369torr \end{gathered}$$