Question

An open flask has Helium gas at 2atm and $327{}^\circ C$. The flask is heated at $527{}^\circ C$ at the same pressure. The fraction of original gas remaining in the flask is?
A. $\dfrac{3}{4}$
B. $\dfrac{1}{4}$
C. $\dfrac{1}{2}$
D. $\dfrac{2}{5}$

Answer 1

Hint: We can assume Helium gas to behave ideally in this case. Therefore, we can use the proportionalities derived from the ideal gas law (as pressure is constant, volume will be directly proportional to temperature) to find the final volume fraction. This is facilitated by the fact that pressure is kept constant in this case.
Formulas used: $\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}$
Where $V$ is the volume, $T$ is the absolute temperature and where the subscripts 1 and 2 indicate the initial and final stages respectively.

Complete step by step solution:
From the ideal gas law, we have: $PV = nRT$
Where $P$ is the pressure of the system, $V$ is the volume, $n$ is the number of moles of the gas, $R$ is the universal gas constant and $T$ is the absolute temperature.
Here as the number of moles is constant, and pressure is constant at $2atm$, $\dfrac{V}{T} = \dfrac{{nR}}{P}$ is a constant and thus, we have the following relation between the initial and final stages:
$\dfrac{{{V_1}}}{{{T_1}}} = dfrac{{{V_2}}}{{{T_2}}}$ where the subscripts 1 and 2 indicate the initial and final stages respectively.
We have to convert the temperature from Celsius to Kelvin scale, by adding 273.
$ \Rightarrow {T_1} = 327 + 273 = 600K,{T_2} = 527 + 273 = 800K$
Substituting the respective values, we get:
$\dfrac{{{V_1}}}{{600}} = \dfrac{{{V_2}}}{{800}}$
Rearranging, we get:
${V_2} = \dfrac{{800{V_1}}}{{600}} = \dfrac{4}{3}{V_1}$
But the flask can only hold a fixed volume, which is equal to $V = {V_1}$, that is, the volume held initially.
Therefore, amount of volume expelled $ = {V_2} - {V_1}$
Substituting the value of ${V_2}$ as obtained above, we get:
amount of volume expelled $ = \dfrac{4}{3}{V_1} - {V_1} = \dfrac{1}{3}{V_1}$
Thus, the fraction of volume expelled is the ratio between the amount of volume expelled and the new volume.
$ \Rightarrow \dfrac{{\dfrac{{{V_1}}}{3}}}{{\dfrac{{4{V_1}}}{3}}} = \dfrac{1}{4}$
Therefore, the volume remaining in the flask is $ = 1 - \dfrac{1}{4} = \dfrac{3}{4}$
Hence, the correct option is option A.

Additional information:
The pressure in this numerical example is kept constant from the beginning of the process till its end. Such a thermodynamic process is known as an isobaric process and is quite common in industries

Note:
Although the volume occupied by the gas changes in this process, note that we are asked to find the fraction of it left in the flask. So, the important thing to note here is that even though the flask is open, it can only hold a fixed amount of gas, which we have mentioned here as ${V_1}$ . Any volume generated more than ${V_1}$ will be expelled out. Note that while computing the fraction, the denominator has to be the new volume. Many times students opt to choose the initial volume of the gas as the denominator, thus leading to a wrong option.