Question

A rigid tank contains 35 kg of nitrogen of 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:
A. 10 kg
B. 12.5 kg
C. 15 kg
D. 17.5 kg

Answer 1

Hint:To solve this question, we need to recall the gas law. According to Robert Boyle, at constant temperature, the pressure P of a gas is inversely proportional to the volume of a gas. It is mathematically represented as PV = k, where k is a constant.

Complete step by step answer:

Now, we will refer to the question which says that the tank contained 35 kg of ${N_2}$ gas at 6 atm. Then, the pressure increases to 9 atm but the temperature remains constant.
So, now to derive the amount of oxygen supplied to the tank we will refer to the gas law.
From gas law we know that,
$PV = nRT$
Where
P= Pressure of the gas
V=Volume
T=Temperature
R=gas constant
$P\infty n \Rightarrow P = kn$
Increase in pressure = $\dfrac{{9 - 6}}{2} = \dfrac{3}{2} = 1.5$
Thus, pressure increases by 1.5 times so the number of moles also increase by 1.5 times.
Now, we know that oxygen has to be added half the number of moles of nitrogen.
So, number of moles of ${N_2}$ in \[35\;kg\; = {\text{ }}143500{\text{ }}\; = {\text{ }}2500\;moles\]
Hence, number of moles of oxygen\[\; = 22500{\text{ }} = {\text{ }}1250\;moles\]
From this, we can say that one mole of oxygen will weigh =$16g$
So, \[1250\;moles\]of oxygen will weight = \[\]\[1250{\text{ }}moles{\text{ }} \times {\text{ }}16{\text{ }}g{\text{ }} = 20000{\text{ }}g = {\text{ }}20{\text{ }}kg\]
Therefore, the amount of oxygen added to increase the pressure from \[6\;atm{\text{ }}to\;9\;atm\] will be $20g$


Note:
We must remember that an ideal gas law states that the product of the pressure and the volume of one gram molecule of an ideal gas is equal to the product of the absolute temperature of the gas and the universal gas constant. The universal gas constant is represented with R which is equal to \[8.3245{\text{ }}Joule{\text{ }}per{\text{ }}mol{\text{ }}per{\text{ }}K\].