- Home
- Standard 11
- Physics
When a system is taken from a state $i$ to $f$ along the path $iaf$ (as shown in the figure). $Q = 50\, cal$ and $W = 20\, cal$ ; along path $ibf,$ $Q = 36\, cal.$
$(i)$ What is $W$ along path $ibf$ ?
$(ii)$ If $W = 13$ cal for path $fi$, what is $Q$ for the path $fi$ ?
$(iii)$ Take $E_{int,i} = 10\,\, cal$ then what is $E_{int,f}$ ?
$30, 20, 40\, cal$
$6, -43, 40\, cal$
$10, -20, 30\, cal$
$15, 35, 25\, cal$
Solution
For path iaf
$Q=50 \mathrm{cal}$
$\mathrm{W}=20 \mathrm{cal}$
According ot I law of thermodynamics,
$\mathrm{dQ}=\mathrm{dU}+\mathrm{d} \mathrm{W}$
$\text { or } \mathrm{dU}=\mathrm{dQ}-\mathrm{dW}=50-20=30 \mathrm{cal}$
(1) For path iaf
$Q=36\,cal$
${W=?}$
$\mathrm{dU}=30 \mathrm{cal}($ since internal energy depends only on the initial and final positions of the system).
$\therefore W=Q-d U=36-30=6 \mathrm{cal}$
$\begin{aligned} \text { (u) } \mathrm{W}=-13 \mathrm{cal} \\ \mathrm{d} \mathrm{U}=-30 \mathrm{cal} \\ \mathrm{Q}=? \end{aligned}$
$\therefore \mathrm{Q}=\mathrm{dU}+\mathrm{W}=-43 \mathrm{cal}$
(iii) $\quad \mathrm{E}_{\text {int }, \mathrm{f}}=\mathrm{E}_{\text {int }, i}+\Delta \mathrm{U}=10 \mathrm{cal}+30 \mathrm{cal} .=40$
cal.
