Question

A cylinder contains 10kg of gas at a pressure of ${{10}^{7}}N/{{m}^{2}}$. The quantity of gas taken out of the cylinder, if final pressure is $2.5\times {{10}^{6}}N/{{m}^{2}}$ is?
A. 9.5 kg
B. 7.5 kg
C. 14.2 kg
D. Zero

Answer 1

We have a gas inside a cylinder. The initial mass, initial pressure and final pressure inside the container is given. We need to find the mass of the gas that is taken out of the container. By using the ideal gas equation, excluding the constant terms we can find a relation between pressure and mass of the gas. With this relation we can solve for the amount of gas taken out of the container.

Formula used:
Ideal gas equation,
$PV=nRT$
Number of moles,
$n=\dfrac{m}{M}$

Complete answer:
We have a cylinder filled with a gas.
The initial mass of the gas is given to be, ${{m}_{i}}=10kg$
And the initial pressure of the gas is also given,
${{P}_{i}}={{10}^{7}}N/{{m}^{2}}$
It is said that a certain amount of the gas is taken out of the container and its final pressure is given.
We have to find the amount of gas taken out.
Final pressure is given to be,
${{P}_{f}}=2.5\times {{10}^{6}}N/{{m}^{2}}$
We know the ideal gas equation,
$PV=nRT$ , were ‘V’ is the volume, ‘P’ is the pressure, ‘R’ is the gas constant, ‘n’ is the total number of moles and ‘T’ is the temperature.
Here we can assume the temperature to be constant and we know ‘R’ is also a constant.
Now let us rewrite the ideal gas equation, we get
$P=\dfrac{n}{V}RT$
Since the container of the gas is the same, we can say that the volume also will be constant.
Therefore we can neglect all the constant terms.
We know number of moles is calculated by the equation,
$\text{n=}\dfrac{\text{given}\text{.mass}}{\text{molar}\text{.mass}}$
$\text{n=}\dfrac{\text{m}}{\text{M}}$, where ‘m’ is the given mass and ‘M’ is the molar mass of the gas.
Now we can rewrite the ideal gas equation as,
$P=\dfrac{m}{M}$
We know that the molar mass of a gas is constant.
Therefore from the above equation we can infer that,
$P\propto m$, i.e. pressure of the gas is directly proportional to the mass of the gas.
From this relation we get,
$\dfrac{{{P}_{f}}}{{{P}_{i}}}=\dfrac{{{m}_{f}}}{{{m}_{i}}}$
From this relation we will get the final mass of the gas,
${{m}_{f}}=\dfrac{{{P}_{f}}{{m}_{i}}}{{{P}_{i}}}$
$\begin{aligned} & {{m}_{f}}=\dfrac{\left( 2.5\times {{10}^{6}} \right)\left( 10 \right)}{{{10}^{7}}} \\\ & {{m}_{f}}=2.5kg \\\ \end{aligned}$
Now we have the final mass of the gas, ${{m}_{f}}=2.5kg$.
We need to find the amount of mass taken out of the cylinder.
This is the difference of initial mass and final mass.
Mass of gas taken out $={{m}_{i}}-{{m}_{f}}=10-2.5=7.5kg$

Hence the correct answer is option B.

Note:
Here we say that the volume of the gas is constant.
When we remove a certain amount of gas from the container, the mass of the gas will get decreased. But the container remains the same, and so does the volume of the container. Therefore the mass of gas left in the container expands and fills the volume of the container.
Hence the volume of the gas remains the same.