Question

The gas in a vessel is subjected to a pressure of 20 atmospheres at a temperature 27$^\circ$C. The pressure of the gas in the vessel after one half of the gas is released from the vessel and the temperature of the remainder is raised by 50$^\circ$C, is
$\begin{array}{l} {\rm{A}}{\rm{. 8}}{\rm{.5 atm}}\\\ {\rm{B}}{\rm{. 10}}{\rm{.8 atm}}\\\ {\rm{C}}{\rm{. 11}}{\rm{.7 atm}}\\\ {\rm{D}}{\rm{. 17 atm}} \end{array}$

Answer 1

Hint: For the given gas, we can apply the ideal gas equation. In this equation, the mass/amount of the gas is specified by the number of moles of the gas. The number of moles decreases with the decrease in the amount of gas.

Formula used:
The ideal gas equation can be given as
$PV = nRT$
where pressure is represented by P, volume is represented by V, n signifies the number of moles of the given gas while T is the temperature of the gas.
R is called the universal gas constant. Its value is given as
R$= 8.314J/mol{\rm{ K}}$
No. of moles of the gas is given as
$n = \dfrac{m}{M}$
where m signifies the available amount of gas and M signifies the molar mass of the gas.

Detailed step by step solution:
We are given gas in a vessel. The pressure of the gas is given as
$P = 20atm$
The temperature of the gas given to be
$T = 27^\circ C = 27 + 273 = 300K$
Also if mass of the gas is taken to be m, the volume to be V then we can write the following ideal gas equation.
$PV = \dfrac{m}{M}RT$
Now the half of the gas is released from the vessel therefore the mass of the gas becomes half and we can write that
$m' = \dfrac{m}{2}$
Also the temperature is raised by 50$^\circ$C, therefore, the final temperature is given as
$T' = 27 + 50 = 77^\circ C = 77 + 273 = 350K$
Now we can write the new equal for the gas as follows:
$P'V = \dfrac{{m'}}{M}RT'{\rm{ }}...{\rm{(ii)}}$
Now we will divide equation (i) by equation (ii). Doing so we get
$\begin{array}{l} \dfrac{P}{{P'}} = \dfrac{m}{{m'}}\dfrac{T}{{T'}}\\\ \Rightarrow P' = \dfrac{{m'}}{m}\dfrac{{T'}}{T}P \end{array}$
Now inserting all the known values, we get
$P' = \dfrac{{\left( {\dfrac{m}{2}} \right)}}{m}\dfrac{{350}}{{300}} \times 20 = \dfrac{{350 \times 20}}{{2 \times 300}} = 11.7atm$
Hence, the correct answer is option C.

Note: 1. The volume occupied by the gas remains the same, only the amount of gas decreases while the dimensions of the container remain the same.
2. An ideal gas has no interactions within its particles. There particles of the gas possess kinetic energy but they do not have any potential energy. Practically, such types of gases do not exist. There are always some interactions between the particles of a gas.