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

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

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