In which process, the net work done is zero?
A. Cyclic
B. Free expansion
C. Isochoric
D. Adiabatic
Answer
633.3k+ views
Hint: Try to recall that work is a path function and it depends only on the path taken by the system not on the initial and final state of the system. Now by using this you can easily find the correct option to the given question.
Formula: \[W = - {P_{ext}}\Delta V\] where W is work done by the system or on the system, Pext is the net external pressure and $\Delta V$ is the change in volume of the system.
Complete Step by step solution:
It is known to you that work done by a system is the amount of energy transferred from the system to its surrounding. There are many kinds of work done out of which P-V work done is used in thermodynamics. For P-V work done, $W = - {P_{ext}}\Delta V$.Now by calculating work done in each of the given options, you can easily find the correct option.
In a cyclic process, the initial and final state of a system is the same but it is known to you that work done does not depend on the initial and final state of the system and hence, is not equal to zero. So, option A is not correct.
In Free expansion, the external pressure is zero and since work done by the system,
$
W = - {P_{ext}}\Delta V \\
\Rightarrow W = (0)\Delta V \\
W = 0 \\
$
So, work done by the system in case of free expansion is 0. Hence, option B is the correct option.
In an isochoric process, volume remains constant i.e. there is no change in volume of the system $(\Delta V = 0)$. So, work done by the system $W = - {P_{ext}}\Delta V$
$
\Rightarrow W = - {P_{ext}}(0) \\
W = 0 \\
$
So, work done by the system in an isochoric process is also 0. Hence the option C is also correct.
In adiabatic process, work done by the system $W = - {P_{ext}}\Delta V$ is not equal to zero as both ${P_{ext}}$ and $\Delta V$ is not equal to 0. So, option C is also not correct.
Therefore, from above we can clearly say that both option B and C are correct options to the given question.
Note: It should be remembered to you that the negative sign in expression of P-V work done indicates that if the volume of the system decreases, then work done by the system is positive and vice-versa.
Also, you should remember that in case of an isobaric process, change in pressure is equal to 0.
Formula: \[W = - {P_{ext}}\Delta V\] where W is work done by the system or on the system, Pext is the net external pressure and $\Delta V$ is the change in volume of the system.
Complete Step by step solution:
It is known to you that work done by a system is the amount of energy transferred from the system to its surrounding. There are many kinds of work done out of which P-V work done is used in thermodynamics. For P-V work done, $W = - {P_{ext}}\Delta V$.Now by calculating work done in each of the given options, you can easily find the correct option.
In a cyclic process, the initial and final state of a system is the same but it is known to you that work done does not depend on the initial and final state of the system and hence, is not equal to zero. So, option A is not correct.
In Free expansion, the external pressure is zero and since work done by the system,
$
W = - {P_{ext}}\Delta V \\
\Rightarrow W = (0)\Delta V \\
W = 0 \\
$
So, work done by the system in case of free expansion is 0. Hence, option B is the correct option.
In an isochoric process, volume remains constant i.e. there is no change in volume of the system $(\Delta V = 0)$. So, work done by the system $W = - {P_{ext}}\Delta V$
$
\Rightarrow W = - {P_{ext}}(0) \\
W = 0 \\
$
So, work done by the system in an isochoric process is also 0. Hence the option C is also correct.
In adiabatic process, work done by the system $W = - {P_{ext}}\Delta V$ is not equal to zero as both ${P_{ext}}$ and $\Delta V$ is not equal to 0. So, option C is also not correct.
Therefore, from above we can clearly say that both option B and C are correct options to the given question.
Note: It should be remembered to you that the negative sign in expression of P-V work done indicates that if the volume of the system decreases, then work done by the system is positive and vice-versa.
Also, you should remember that in case of an isobaric process, change in pressure is equal to 0.
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