NEET-XII-Physics
26: Laws of Thermodynamics
- #4Refer to figure. Let â–µU1 and â–µU2 be the change in internal energy in processes A and B respectively, â–µQ be the net heat given to the system in process A + B and â–µW be the net work done by the system in the process A + B.
Figure
(a) â–µU1 + â–µU2 = 0.
(b) â–µU1 - â–µU2 = 0.
(c) â–µQ - â–µW = 0.
(d) â–µQ + â–µW = 0digAnsr: a,cAns :
(a) ​∆U1 + ∆U2 = 0
(c) ∆Q - ∆W = 0
The process that takes place through A and returns back to the same state through B is cyclic. Being a state function, net change in internal energy, ∆U will be zero, i.e.
∆U1 (Change in internal energy in process A) = ∆​U2 (Change in internal energy in process B)
`` \Rightarrow \Delta U=\Delta {U}_{1}+\Delta {U}_{2}=0``
Here, ∆U is the total change in internal energy in the cyclic process.
Using the first law of thermodynamics, we get
`` \Delta Q-\Delta W=\Delta U``
Here, ∆Q is the net heat given to the system in process A + B and ∆W is the net work done by the system in the process A + B.
Thus,
∆Q - ∆W = 0
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