Question

The total pressure of a mixture of 8g of oxygen and 14g of nitrogen contained in a 11.2l of vessel at 0 degree centigrade is

Answer 1

Answer:

8 g of oxygen =8/32=0.25 mole.

14 g of nitrogen=14/28=0.5 mole.

So there is a total of (0.25+0.5)=0.75 mole.

At STP 0.75 mole gas will occupy=22.4 x 0.75= 16.8 liters.

At constant temperature PV=P°V°

So P=P°V°/V= 1 x 16.8/11.2=1.5 atm.

Answer 2

Answer:

P = 1.5 atm

Explanation:

The gas equation is given as;

PV = n RT

Now let us calculate the total number of moles of gas mixture.

We are given 8 g of oxygen = 8/32 = 0.25 mole.

And

14 g of nitrogen =14/28=0.5 mole.

Hence value of n = 0.25+0.5 = 0.75 mole

Now applying the formula;

PV = n RT

R = 0.821, T = 273K, V = 11.2

P = 0.75 × 0.821 × 273/11.2

P = 1.5 atm