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Following figure shows two processes $A$ and $B$ for a gas. If $\Delta Q_A$ and $\Delta Q_B$ are the amount of heat absorbed by the system in two cases, and $\Delta U_A$ and $\Delta U_B$ are changes in internal energies, respectively, then
$\Delta {Q_A}\, = \,\Delta {Q_B}\,;\,\Delta {U_A} = \Delta {U_B}$
$\Delta {Q_A}\, > \,\Delta {Q_B}\,;\,\Delta {U_A} = \Delta {U_B}$
$\Delta {Q_A}\, < \,\Delta {Q_B}\,;\,\Delta {U_A} < \Delta {U_B}$
$\Delta {Q_A}\, > \,\Delta {Q_B}\,;\,\Delta {U_A} > \Delta {U_B}$
Solution
Initial and final states for both the processes are same,
$\therefore \Delta \mathrm{U}_{\mathrm{A}}=\Delta \mathrm{U}_{\mathrm{B}}$
Work done during process $A$ is greater than in process $B$. Because area is more
By First law of thermodynamics
$\Delta Q=\Delta U+W$
$\Rightarrow \Delta \mathrm{Q}_{\mathrm{A}}>\Delta \mathrm{Q}_{\mathrm{B}}$
Similar Questions
In Column $-I$ processes and in Column $-II$ formulas of work are given. Match them appropriately :
| Column $-I$ | Column $-II$ |
| $(a)$ Isothermal process | $(i)$ $W = \frac{{\mu R({T_1} – {T_2})}}{{\gamma – 1}}$ |
| $(b)$ Adiabatic process | $(ii)$ $W = P\Delta V$ |
| $(iii)$ $W = 2.303\,\mu RT\log \left( {\frac{{{V_2}}}{{{V_1}}}} \right)$ |
