Question

Two moles of an ideal gas at $300K$ were cooled at constant volume so that the pressure is reduced to half the initial value. Then as a result of heating at constant pressure, the gas expands till it attains the original temperature. Find the total heat absorbed by the gas, if $R$ is the gas constant.
(A) $150R\,J$
(B) $300R\,J$
(C) $75R\,J$
(D) $100R\,J$

Answer 1

Hint: To solve this question we should know about the thermodynamic processes and the relation between them. There are a lot of equations and formulas regarding the thermodynamic processes revising those formulas before solving this sum helps in better understanding.

Complete step by step answer
There are different types of thermodynamic processes.
A process in which the temperature of the system is maintained throughout is called an isothermal process.
In isobaric processes the pressure is maintained constant while in isochoric processes the volume is maintained constant.
If the system is insulated from the surroundings then no heat flows between the system and the surroundings and this process is adiabatic process.
In the given problem the first process is an isochoric process because the volume is constant in the first process.
In isochoric process the heat change is given by
$\Delta {Q_v} = n{C_v}\Delta T$
Where,
$\Delta {Q_V}$ is the change in heat at constant volume
${C_v}$ is the specific heat capacity at constant volume.
$\Delta T$ is the change in temperature.
$n$ is the number of moles.
Given that,
The initial temperature ${T_A} = 300K$
The number of moles $n = 2$
The pressure is reduced to half $P = \dfrac{P}{2}$
In an isochoric process, volume remains constant, so pressure is directly proportional to temperature
The final temperature ${T_B} = \dfrac{{{T_A}}}{2}$
$ \Rightarrow {T_B} = \dfrac{{300}}{2}$
$ \Rightarrow {T_B} = 150K$
The change in temperature is ${T_B} - {T_A} = 300 - 150$
$ \Rightarrow \Delta T = 130K$
The second process is isobaric process because the pressure is constant
In isobaric process the heat change is given by
$\Delta {Q_P} = n{C_p}\Delta T$
Where,
$\Delta {Q_P}$ is the change in heat at constant pressure
${C_P}$ is the specific heat capacity at constant pressure.
$\Delta T$ is the change in temperature.
$n$ is the number of moles.
Net change in heat i.e. the heat absorbed by the gas is given by
$ \Rightarrow \Delta {Q_p} - \Delta {Q_V}{\text{ = }}n{C_p}\Delta T - n{C_v}\Delta T{\text{ = n}}\Delta {\text{T(}}{{\text{C}}_p} - {C_V})$
$ \Rightarrow \Delta {Q_p} - \Delta {Q_V}{\text{ = n}}\Delta {\text{T(}}{{\text{C}}_p} - {C_V})$
In a thermodynamic process, ${\text{(}}{{\text{C}}_p} - {C_V}) = R$, where R is gas constant
$ \Rightarrow \Delta {Q_p} - \Delta {Q_V}{\text{ = n}}\Delta {\text{TR}}$
Substitute the known values
$ \Rightarrow \Delta {Q_p} - \Delta {Q_V}{\text{ = 2}} \times {\text{150}} \times {\text{R}}$
$ \Rightarrow \Delta {Q_p} - \Delta {Q_V}{\text{ = 300R J}}$
The unit of heat is Joule

Hence the correct answer is option (B) $300R\,J$

Note: Most of the students do not know how to find the temperature change in this problem. In an isochoric process, volume remains constant, so pressure is directly proportional to temperature. Here the pressure is reduced to half so the temperature is also reduced to half.