Question

A gas is filled in a container at any temperature and at pressure \[76cm\] of Hg. If at the same temperature the mass of gas is increased 50%, then the resultant pressure will be:
\[\begin{align}
  & \text{A}\text{. 38cm of Hg} \\
 & \text{B}\text{. 76cm of Hg} \\
 & \text{C}\text{.114 cm of Hg} \\
 & \text{D}\text{. 152 cm of Hg} \\
\end{align}\]

Answer 1

Hint: To calculate the pressure when the mass is increased by 50 %, we need to use the ideal gas law. Here, all the necessary quantities are given. So taking the ratio of the system before and after expansion, gives us the required quantity.

Complete step by step answer:
Assuming the given gas is ideal, we can use the ideal gas law .i.e. $PV=nRT$, where $P$ is the pressure of the gas, $T$ is the temperature of the gas, $V$ is the volume of the gas, $n$ is the number of moles of the gas molecule present in the container and $R$ is the gas constant, with value $8.314 KJmol^{-1}K^{-1}$.
Given $T_{1}=T_{2}=T$ , $m_{1}=m$, $m_{2}=\dfrac{m}{2}+m$ and $P_{1}=76cm$ of Hg, then we need to find the $P_{2}$
Here, mass of the gas $m$ is given instead of the number of moles of the gas molecules$n$. But we know that, $n=\dfrac{m}{M}$ where$M$ is the molar mass of the gas.
Then, the gas law becomes, $PV=\dfrac{m}{M}RT$
Then, taking the ratio we get, $\dfrac{P_{1}V_{1}}{P_{2}V_{2}}=\dfrac{\dfrac{m_{1}}{M_{1}}RT_{1}}{\dfrac{m_{2}}{M_{2}}RT_{2}}$, since $T_{1}=T_{2}=T$, $V_{1}=V_{2}=V$ and $M_{1}=M_{2}=M$ and $R$ is a constant, we get,$\dfrac{P_{1}}{P_{2}}=\dfrac{m_{1}}{m_{2}}$
Substituting the values, we get, $P_{2}=\dfrac{m_{2}}{m_{1}}P_{1}=\dfrac{m+\dfrac{m}{2}}{m}\times 76=\dfrac{3m}{2m}\times 76=114cm$of Hg.

Thus the answer is \[\text{C}\text{.114 cm of Hg}\]

Note:
Ideal gas law is the combination of Boyle’s law, Charles’s law and Gay-Lussac’s laws. The gas is true only for ideal gases, hence to apply the formula, we must assume the gas to be ideal. We must also know that the molar mass is a constant, and clearly it is independent of the mass of the gas. Also note that $R$ is the gas constant with value $8.314 KJmol^{-1}K^{-1}$. Also there are other values of the gas constant, depending on the units used.