An ideal gas goes from state A to state B via three different...

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 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

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

Options

A $Q_1>Q_2>Q_3$ and $\Delta U_1=\Delta U_2=\Delta U_3$
B $Q_3>Q_2>Q_1$ and $\Delta U_1=\Delta U_2=\Delta U_3$
C $Q_1=Q_2=Q_3$ and $\Delta U_1>\Delta U_2>\Delta U_3$
D $Q_3>Q_2>Q_1$ and $\Delta U_1>\Delta U_2>\Delta U_3$

Correct answer

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

Step-by-step solution

For all processes 1,2 and 3 Change in internal energy, i.e.

\begin{gathered} \Delta U=U_B-U_A \\ \therefore \Delta U_1=\Delta U_2=\Delta U_3 \Rightarrow Q=\Delta U+\Delta W \end{gathered}

Now,

\Delta W=

work done by the gas, i.e. area under

p-V

curve. As, area

(1)>

area

(2)>

area

(3)

\begin{array}{ll} \therefore & \Delta W_1>\Delta W_2>\Delta W_3 \\ \therefore & Q_1>Q_2>Q_3 \end{array}

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