Question

A thermodynamic system goes from states $(i){P_1},V{\text{ to 2}}{{\text{P}}_1},V{\text{ }}(ii){P_1},{V_1}{\text{ to }}{{\text{P}}_1},2{V_1}$. Then the work done in two cases is respectively
$
  (a){\text{ 0,0}} \\
  (b){\text{ 0,}}{{\text{P}}_1}{V_1} \\
  (c){\text{ }}{{\text{P}}_1}{V_1}{\text{,0}} \\
  (d){\text{ }}{{\text{P}}_1}{V_1}{\text{,}}{{\text{P}}_1}{V_1} \\
 $

Answer 1

- Hint – In this question use the concept that work done is given as \[W = \int_{{V_1}}^{{V_2}} {PdV} \], so in the first case Volume remains the same as the system is going from V to V, and in the second case pressure remains constant and the volume is going from ${V_1}{\text{ to 2}}{{\text{V}}_1}$. Substitute the values in the direct formula to get the work done.

Complete step-by-step solution -

The work done from first state to second state is given as
\[W = \int_{{V_1}}^{{V_2}} {PdV} \], where P is constant w.r.t. V and P and V are the pressure and volume respectively.
Case – 1 P1, V to 2P1, V
Now as we see that pressure is changing and volume remains constant.
\[ \Rightarrow {W_1} = \int_V^V {PdV} = 0\] (as the upper and lower limit is the same so the integration value is zero).
So the work done in this case is zero.
Case – 2 P1, V1 to P1, 2V1
Now as we see in this case pressure remains same but volume changes so the work done is
\[ \Rightarrow {W_2} = \int_{{V_1}}^{2{V_1}} {{P_1}dV} \]
Now integrate it we have,
\[ \Rightarrow {W_2} = {P_1}\left[ V \right]_{{V_1}}^{2{V_1}}\]
Now apply integrating limit we have,
\[ \Rightarrow {W_2} = {P_1}\left[ {2{V_1} - {V_1}} \right] = {P_1}{V_1}\]
So this is the required answer.
Hence option (B) is correct.

Note – In an isochoric process also known as constant-volume process the volume of the closed system undergoing pressure remains constant, since the volume is held constant therefore the work done by the system will always be zero, it follows that for the simple system of two dimensions any energy transferred to the system externally will be absorbed as internal energy however it is not the case with the second part that’s why it’s volume is changing in 2nd part.