JEE MAIN - Physics (2023 - 25th January Evening Shift - No. 4)
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Isothermal Process | I. | Work done by the gas decreases internal energy |
| B. | Adiabatic Process | II. | No change in internal energy |
| C. | Isochoric Process | III. | The heat absorbed goes partly to increase internal energy and partly to do work |
| D. | Isobaric Process | IV. | No work is done on or by the gas |
Choose the correct answer from the options given below :
A-I, B-II, C-IV, D-III
A-II, B-I, C-III, D-IV
A-II, B-I, C-IV, D-III
A-I, B-II, C-III, D-IV
Explanation
$$
\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}
$$
For isothermal process $\mathrm{T}$ is constant
So $\Delta \mathrm{U}=0$
$\mathrm{A} \longrightarrow \mathrm{II}$
Adiabatic process
$$ \begin{aligned} & \Delta \mathrm{Q}=0 \\\\ & \Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W} \\\\ & \Delta \mathrm{U}=-\Delta \mathrm{W} \end{aligned} $$
Work done by gas is positive
So $\Delta \mathrm{U}$ is negative
$$ \text { B } \longrightarrow \text { I } $$
For Isochoric process $\Delta \mathrm{W}=0$
$$ \mathrm{C} \longrightarrow \mathrm{IV} $$
For Isobaric process
$$ \begin{aligned} & \Delta \mathrm{W}=\mathrm{P} \Delta \mathrm{V} \neq 0 \\\\ & \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T} \neq 0 \end{aligned} $$
Heat absorbed goes partly to increase internal energy and partly to do work.
For isothermal process $\mathrm{T}$ is constant
So $\Delta \mathrm{U}=0$
$\mathrm{A} \longrightarrow \mathrm{II}$
Adiabatic process
$$ \begin{aligned} & \Delta \mathrm{Q}=0 \\\\ & \Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W} \\\\ & \Delta \mathrm{U}=-\Delta \mathrm{W} \end{aligned} $$
Work done by gas is positive
So $\Delta \mathrm{U}$ is negative
$$ \text { B } \longrightarrow \text { I } $$
For Isochoric process $\Delta \mathrm{W}=0$
$$ \mathrm{C} \longrightarrow \mathrm{IV} $$
For Isobaric process
$$ \begin{aligned} & \Delta \mathrm{W}=\mathrm{P} \Delta \mathrm{V} \neq 0 \\\\ & \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T} \neq 0 \end{aligned} $$
Heat absorbed goes partly to increase internal energy and partly to do work.
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