Question

Three moles of an ideal gas ${{C}_{P}}=\dfrac{7R}{2}$at pressure​ ${{P}_{A}}$and temperature ​is ${{T}_{A}}$isothermally expanded to twice its initial volume, it is then compressed at constant pressure to its original volume. Finally gas is heated at constant volume to its original pressure${{P}_{A}}$.Calculate the net work done by the gas, and net heat supplied to the gas during the complete process.
$A)W=0.58R{{T}_{A}};Q=1.58R{{T}_{A}}$
$B)W=0.58R{{T}_{A}};Q=0.58R{{T}_{A}}$
$C)W=1.58R{{T}_{A}};Q=0.58R{{T}_{A}}$
$D)W=10.58R{{T}_{A}};Q=10.58R{{T}_{A}}$

Answer 1

First Process is isothermal; pressure is inversely proportional to volume. Since volume is double so pressure is half. Second process is an isobaric process i.e. pressure is constant and changes to initial volume. Third process is an isochoric process i.e. volume is constant and gas reaches initial pressure. The whole system reaches the initial state i.e.$\Delta U=0$.

Complete answer:
Let us assume the number of moles is $n=3$
Specific Heat at the Constant Pressure is given ${{C}_{p}}=\dfrac{7R}{2}$
Let us assume the initial volume is ${{V}_{1}}$ and the final volume is${{V}_{2}}$.
Let us assume the final pressure is${{P}_{2}}$.
According to the Question, the given process is isothermal and volume becomes double of its initial volume.
$\Rightarrow {{V}_{2}}=2{{V}_{1}}$
Since, Process is Isothermal i.e. Temperature is constant so Pressure and Volume are inversely proportional to each other.
$PV=Cons\tan t$.
\Rightarrow $$$${{P}_{A}}{{V}_{1}}={{P}_{2}}{{V}_{2}}
Final pressure will become
\Rightarrow {{P}_{2}}=\dfrac{{{P}_{A}}{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{P}_{A}}}{2}$$$$(Equation1)
${{W}_{iosthermal}}=nR{{T}_{A}}\ln \dfrac{{{P}_{i}}}{{{P}_{f}}}$
${{W}_{isothermal}}=3R{{T}_{A}}\ln \dfrac{{{P}_{A}}}{{{P}_{2}}}$
From Equation 1 Put the value of PA
${{W}_{isothermal}}=3R{{T}_{A}}\ln \dfrac{2{{P}_{2}}}{{{P}_{2}}}$
${{W}_{isothermal}}=3R{{T}_{A}}\ln 2$ $(Equation2)$
So final Pressure becomes P2, now it is compressed to its initial volume at same constant pressure.
So work done at constant pressure will become${{W}_{1}}$.
${{W}_{1}}={{P}_{2}}({{V}_{f}}-{{V}_{i}})$
${{W}_{1}}=\dfrac{{{P}_{A}}}{2}({{V}_{2}}-{{V}_{A}})$
${{W}_{1}}=\dfrac{{{P}_{A}}}{2}(2{{V}_{1}}-{{V}_{1}})$
${{W}_{1}}=\dfrac{{{P}_{A}}{{V}_{1}}}{2}$
Since it is a compression so work done is taken as negative.
{{W}_{1}}=-\dfrac{{{P}_{A}}{{V}_{1}}}{2}$$$$(Equation3)
Now gas is heated at constant volume so no work is done.
Let us assume work done for this state is${{W}_{2}}$.
So,${{W}_{2}}=0$.$(Equation4)$
So Total work done is = ${{W}_{isothermal}}+{{W}_{1}}+{{W}_{2}}$
Put the values of work done from Equation 2 , Equation 3 and Equation 4 respectively.
We get ,
{{W}_{total}}=3R{{T}_{A}}\ln 2+(\dfrac{-{{P}_{A}}{{V}_{1}}}{2})+0$$$$(Equation5)
According to ideal gas equation $PV=nRT$
we get, ${{P}_{A}}{{V}_{1}}=3R{{T}_{A}}$.
Put this value in Equation 5
${{W}_{total}}=3R{{T}_{A}}\ln 2-\dfrac{3R{{T}_{A}}}{2}$
${{W}_{total}}=3R{{T}_{A}}(\ln 2-\dfrac{1}{2})$
${{W}_{total}}=3R{{T}_{A}}(0.693-0.5)$
${{W}_{total}}=3R{{T}_{A}}(0.193)$
${{W}_{total}}=0.579R{{T}_{A}}$
${{W}_{total}}\approx 0.58R{{T}_{A}}$
Now, Let us calculate heat in each process.
Since after travelling from several states gas reaches its initial state. So no change in internal energy of the gas
$\Rightarrow \Delta U=0$
So According to first law of thermodynamics ,
$\Delta Q=\Delta U+\Delta W$
$\Rightarrow \Delta Q=\Delta W$
$\Rightarrow {{Q}_{total}}={{W}_{total}}$
so total heat energy supplied is given by ,
${{Q}_{total}}=0.58R{{T}_{A}}$
So total work done by the gas is ${{W}_{total}}\approx 0.58R{{T}_{A}}$and net heat supplied to the gas during the complete process is ${{Q}_{total}}=0.58R{{T}_{A}}$

Correct option is B.

Note:
During compression work done is taken as a negative sign while during expansion work done is taken as a positive sign. When the system reaches its initial state then change in internal energy is zero. Then according to the first law of thermodynamics heat energy of gas is equal to work done by the system.