Heat and Thermodynamics 6 Question 13
17. Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be
(2014 Adv.)
(a) $250 R$
(b) $200 R$
(c) $100 R$
(d) $-100 R$
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Answer:
Correct Answer: 17. (d)
Solution:
- $\Delta W _1+\Delta U _1=\Delta Q _1\cdots(i)$
$\Delta W _2+\Delta U _2=\Delta Q _2\cdots(ii)$
$\Delta Q _1+\Delta Q _2=0$
$\therefore\left(n C _p \Delta T\right) _1+\left(n C _p \Delta T\right) _2=0$
But $\quad n _1=n _2=2$
$\therefore \quad \frac{5}{2} R(T-700)+\frac{7}{2} R(T-400)=0$
Solving, we get $T=525 K$
Now, from Eqs. (i) and (ii), we get
$$ \begin{array}{cc} \Delta W _1+\Delta W _2=-\Delta U _1-\Delta U _2 \\ \text { as } & \Delta Q _1+\Delta Q _2=0 \\ \therefore & \Delta W _1+\Delta W _2=-\left[\left(n C _V \Delta T\right) _1+\left(n C _V \Delta T\right) _2\right] \\ =-\left[2 \times \frac{3}{2} R \times(525-700)+2 \times \frac{5}{2} R \times(525-400)\right] \\ =-100 R \end{array} $$
