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In the $P-V$ diagram, $I$ is the initial state and $F$ is the final state. The gas goes from $I$ to $F$ by $(i)\,IAF,\,\,(ii)\,IBF,\,\,(iii) ICF$. The heat absorbed by the gas is
the same in all three processes
the same in $(i)$ and $(ii)$
greater in $(i)$ than in $(ii)$
the same in $(i)$ and $(iii)$
Solution
Heat absrobed in a thermodynamic process is given by
$\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$
Here, $\Delta \mathrm{U}$ is same for all the three processes as it depends only on initial and final states.
$\text { But } \quad \Delta \mathrm{W}_{1}=+\mathrm{ve}, \Delta \mathrm{W}_{\mathrm{II}}=0$
${\text { and }} {\Delta \mathrm{W}_{\mathrm{III}}=-\mathrm{ve}} $
${\text { or }} {\Delta \mathrm{Q}_{1}>\Delta \mathrm{Q}_{\mathrm{II}}}$
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 | $(ii)$ $W = P\Delta V$ |
| $(iii)$ $W = 2.303\,\mu RT\log \left( {\frac{{{V_2}}}{{{V_1}}}} \right)$ |
