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  • Figure shows the P − V diagram for an ideal gas. From the graph we conclude that <img alt="" src="https://cdn.entrance360.com/media/uploads/2023/08/19/phy-f56-q72.PNG" style="height

Figure shows the P − V diagram for an ideal gas. From the graph we conclude that

Option: 1

The process A → B is adiabatic


Option: 2

The internal energy of the gas increases in this process


Option: 3

The work done by the gas = area of triangle ABC


Option: 4

The heat energy absorbed by the gas is zero in the process


Answers (1)

best_answer

The P − V graph for an adiabatic process is not a straight-line. Hence choice (a) is
wrong.\mathrm{P_A V_A=n R T_A \text { and } P_B V_B=n R T_B}. Therefore

\mathrm{\frac{P_A V_A}{P_B V_B}=\frac{T_A}{T_B} \Rightarrow \frac{6 \times 1}{2 \times 4} \Rightarrow \frac{T_A}{T_B}=\frac{3}{4}}

\mathrm{\text { i.e. } T_B>T_A \text {. }} TA. Hence the internal energy increases.
Work done = area under AB upto the volume axis. Heat energy is absorbed in the
process

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Gunjita

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