Question 65: Thermodynamics & KTG

Question

An ideal gas goes from state $A$ to state $B$ via three different processes as indicated in the $p$-$V$ diagram. If $Q_1, Q_2$ and $Q_3$ indicate the heat absorbed by the three processes and $\Delta U_1, \Delta U_2$ and $\Delta U_3$ indicate the change in internal energy along the three processes respectively, then
![](https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lqbz0y15/287e5b6e-5211-4160-b867-86d4d9374f26/d6c31390-9e38-11ee-88dd-593bb30ed511/file-6y3zli1lqbz0y16.png?format=png)

Accepted Answer

For all processes 1, 2 and 3
Change in internal energy, i.e.
$\begin{aligned}
& \quad \Delta U=U_B-U_A \
& 	herefore \Delta U_1=\Delta U_2=\Delta U_3\Rightarrow Q=\Delta U+\Delta W
\end{aligned}$
Now, $\Delta W=$ work done by the gas, i.e. area under $p$-$V$ curve.
$\begin{aligned}
& 	ext { As, area }(1) > \operatorname{area}(2) > 	ext { area }(3) \
& 	herefore \quad \Delta W_1 > \Delta W_2 > \Delta W_3 \
& 	herefore \quad Q_1 > Q_2 > Q_3
\end{aligned}$

Answer

$Q_1 > Q_2 > Q_3$ and $\Delta U_1=\Delta U_2=\Delta U_3$

Answer

$Q_3 > Q_2 > Q_1$ and $\Delta U_1=\Delta U_2=\Delta U_3$

Answer

$Q_1=Q_2=Q_3$ and $\Delta U_1 > \Delta U_2 > \Delta U_3$

Answer

$Q_3 > Q_2 > Q_1$ and $\Delta U_1 > \Delta U_2 > \Delta U_3$

Thermodynamics & KTG

An ideal gas goes from state A to state B via three different processes as indicated in the p-V diagram. If Q_1, Q_2 and Q_3 indicate the heat absorbed by the three processes and \Delta U_1, \Delta U_2 and \Delta U_3 indicate the change in internal energy along the three processes respectively, then

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