Q. 8.78

Question

A mixture of nitrogen $(N_{2})$ and helium has a volume of $250 m L$ at $30 ° C$ and a total pressure of $745 m m H g$ .

a. If the partial pressure of helium is $32 m m H g$ , what is the partial pressure of the nitrogen?

b. What is the final volume, in litres, of the nitrogen at STP?

Step-by-Step Solution

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Answer

a. The partial pressure of the nitrogen gas is $713 m m H g$ .

b. The final volume of the nitrogen is $0.211 L$ .

1Part (a) Step 1: Given Information

We need to find the partial pressure of the nitrogen with the pressure of helium is $32 m m H g$ given.

2Part (a) Step 2: Simplify

Dalton's law of partial pressure says that the total pressure of the gas is equal to the sum of the individual pressure of gasses present in a container.

$P_{t o t a l} = P_{H e} + P_{N_{2}} . . . (1)$

Considering the given values

The total pressure $P_{t o t a l} = 745 m m H g$ and

the pressure of helium $P_{H e} = 32 m m H g$

Now, calculate the pressure of nitrogen using the given values. So the equation $(1)$ will be

$745 m m H g = 32 m m H g + P_{N_{2}} P_{N_{2}} = 745 m m H g - 32 m m H g P_{N_{2}} = 713 m m H g$

3Part (b) Step 1: Given Information

We need to find the final volume of the nitrogen.

4Part (b) Step 2: Simplify

Considering the given values

pressure of nitrogen $P_{N} = 1.0 a t m$ ,

temperature of nitrogen $T_{N} = 273.15 K$ , and

the total pressure $P_{T} = 745 m m H g = \frac{745}{760} a t m = 0.98026 a t m$ ,

the total volume $V_{T} = 250 m L$ , and

the total temperature $T_{T} = 30 ° C = (30 + 273) K = 303 K$ .

First, calculating the total number of moles of the gas using the ideal gas law

$P_{T} V_{T} = n_{T} R T_{T} n_{T} = \frac{P_{T} V_{T}}{R T_{T}} n_{T} = \frac{0.98026 a t m \times 0.250 L}{0.0821 \frac{a t m - L}{m o l - K} \times 303.15 K} n_{T} = 0.00986 m o l$