Question

TA vessel has 6g of oxygen at pressure P and temperature 400K.A small hole is made in it so that oxygen leaks out. How much oxygen leaks out if the final pressure is P/2 and temperature is 300K?
A. 3g
B. 2g
C. 4g
D. 5g

Answer 1

Hint: In this question we can apply an ideal equation to calculate the quantity of oxygen leaked. Ideal gas equations can be applied for any ideal gas. By taking the difference of two masses we can easily get the quantity of leaked oxygen.

Complete step by step solution:
First let us discuss about gaseous:
There are two types of gaseous real and ideal gaseous. At low pressure and high temperature gases behave ideally.
We have ideal gas equation:
$PV = nRT$
Where P-pressure of the gas, V-volume of the gas, n-number of moles of the gas, R- gas constant, T-temperature
Given: Initial mass of oxygen () = 6g, Initial pressure of gas = P, Temperature (T) = 400K
After gas leaked out, Final Pressure () =, Temperature () = 300K
Now apply ideal gas equation,
$PV = nRT$
\[number{\text{ }}of{\text{ }}moles{\text{ }} = {\text{ }}mass{\text{ }}of{\text{ }}the{\text{ }}gas/Molar{\text{ }}mass{\text{ }}of{\text{ }}the{\text{ }}gas\]
$P \times V = \dfrac{6}{{32}} \times R \times 400 - (i)$
$\dfrac{P}{2} \times V = \dfrac{{m'}}{{32}} \times R \times 300 - (ii)$
Dividing both equations, we get
$m' = 4g$
So leaked mass is 2g.

Hence our correct option is B.

Note:
In similar types of problems you can directly use the ideal gas equation. Also remember that the mass came on solving which is 4g is remaining mass not the mass leaked so don’t forget to subtract to get the leaked mass.