Heat And Thermodynamics Ques 171
- Following figure shows two processes $A$ and $B$ for a gas. If $\Delta Q _A$ and $\Delta Q _B$ are the amount of heat absorbed by the system in two cases, and $\Delta U _A$ and $\Delta U _B$ are changes in internal energies respectively, then
(2019 Main, 9 April I)
(a) $\Delta Q _A>\Delta Q _B, \Delta U _A>\Delta U _B$
(b) $\Delta Q _A<\Delta Q _B, \Delta U _A<\Delta U _B$
(c) $\Delta Q _A>\Delta Q _B, \Delta U _A=\Delta U _B$
(d) $\Delta Q _A=\Delta Q _B ; \Delta U _A=\Delta U _B$
Show Answer
Answer:
Correct Answer: 171.(c)
Solution:
Formula:
- According to the first law of thermodynamics,
Heat supplied $(\Delta Q)=$ Work done $(W)+$ Change in internal energy of the system $(\Delta U)$
Similarly, for process $B$,
$$ \Delta Q _A=\Delta U _A+W _A $$
Now, we know that,
$$ \Delta Q _B=\Delta U _B+W _B $$
work done for a process $=$ area under it’s $p$ - $V$ curve Here,
Thus, it is clear from the above graphs,
$$ W _A>W _B \cdots(i) $$
Also, since the initial and final state are same in both process, so
$$ \Delta U _A=\Delta U _B \cdots(ii) $$
So, from Eqs. (i) and (ii), we can conclude that
$$ \Delta Q _A>\Delta Q _B $$
BETA
