Question

A rigid tank contains 35 kg of nitrogen at 6 atm. Sufficient quantity of oxygen is supplied to increase the pressure to 9 atm, while the temperature remains constant. Amount of oxygen supplied to the tank is

Answer 1

The amount of oxygen added to increase the pressure from 6 atm to 9 atm will be 20 kg.

From the gas law we know that:

PV = nRT

since V,R and T are constant here

=> P ∝ n

since pressure is increased by 1.5 times, hence number of moles should also increase by 1.5 times

so oxygen added should be equal to half the number of moles of nitrogen.

Number of moles of nitrogen in 35 kg = 35000/14 = 2500 moles

Hence number of moles of oxygen = 2500/2 = 1250 moles

One mole of oxygen weight = 16 g

=> 1250 moles of oxygen weight = 1250 x 16 = 20000 g = 20 kg

Hence the amount of oxygen added to increase the pressure from 6 atm to 9 atm will be 20 kg.

Answer 2

A rigid tank contains 35kg of nitrogen at 6atm.

so, number of mole of nitrogen = given weight/atomic weight
= 35000g/14g/mol
= 2500 moles

here you should notice that, temperature and volume do not change .

from gas equation, PV = nRT ,
V, R, T are constant terms then, P$\propto$ n

e.g., pressure is directly proportional to number of mole.

here pressure increases 6atm to 9atm. e.g., pressure becomes 1.5 times of initial. so, final number of moles will become 1.5 times.

so, number of mole = 1.5 × 2500 = 37500
but 2500 moles is nitrogen in 37500 moles
so, number of mole of oxygen = 37500 - 2500
= 12500

now, number of mole of oxygen = mass of oxygen/atomic mass of oxygen

12500 = mass of oxygen/16

mass of oxygen = 16 × 12500 = 20,000g

hence, mass of oxygen = 20kg.