Question

A flask is filled with \[13{\text{ gm}}\] of an ideal gas at a temperature of ${27^ \circ }C$ . Its temperature is raised to \[{52^ \circ }C\] . With pressure remaining the same, what is the mass of the gas that should be released to maintain the temperature of the gas in the flask at ${52^ \circ }C$?
A. \[{\text{2}}{\text{.5 gm}}\]
B. \[{\text{2}}{\text{.0 gm}}\]
C. \[1.5{\text{ gm}}\]
D. $1.0{\text{ gm}}$

Answer 1

Hint:According to ideal gas law, $PV = nRT$ , where $P$ is the pressure of the gas, $V$ is its volume, $n$ is the number of moles, $T$ is temperature and $R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ . Use the ideal gas law to solve the given question.

Formula used:
Ideal gas law,
$PV = nRT$
Here, $P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles,
$T$ is the temperature and
$R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ .

Complete answer:
According to ideal gas law,
$PV = nRT$ … (1)
It is given that a flask is filled with \[13{\text{ gm}}\] of an ideal gas at a temperature of ${27^ \circ }C{\text{ }}(300{\text{ K}})$ .
At this condition, let the number of moles of the gas be ${n_1}$.
As the volume of the gas is equivalent to the volume of the flask and as it is given that the pressure is same, therefore, let these values of volume and pressure be $V$ and $P$ respectively.
Also, let \[{m_1} = 13{\text{ gm}}\] and ${T_1} = 300{\text{ K}}$ .
Applying ideal gas law as given in formula (1), we get:
${n_1} = \dfrac{{PV}}{{300R}}$ … (2)
The temperature was then raised to \[{52^ \circ }C{\text{ }}(325{\text{ K}})\] .
At this condition, let the number of moles of the gas be ${n_2}$and let the mass of the gas be \[{m_2}\] .
Also, let ${T_2} = 325{\text{ K}}$ .
Applying ideal gas law as given in formula (1) for this condition, we get:
${n_2} = \dfrac{{PV}}{{325R}}$ … (3)
Dividing equation (2) by (3),
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{325}}{{300}} = \dfrac{{13}}{{12}}$ … (4)
Now, we know that the ratio of the moles of the gas will give us the ratio of the mass of the gas. Thus,
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{{m_1}}}{{{m_2}}}$ … (5)
Substituting the value of $\dfrac{{{n_1}}}{{{n_2}}}$ from equation (4) in the above equation,
$\dfrac{{13}}{{{m_2}}} = \dfrac{{13}}{{12}}$
On simplifying, ${m_2} = 12{\text{ gm}}$ .
Mass of the gas that should be released $ = {m_2} - {m_1} = 13 - 12 = 1{\text{ gm}}$
Thus, the correct option is D.



Note: In the given question, the volume and pressure of the gas remain unchanged on increasing the temperature. We also know that the ratio of the moles of the gas will give us the ratio of the mass of the gas. Thus, you could have also used ${m_1}{T_1} = {m_2}{T_2}$ to get the required mass, as all the other values are constant.