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**Hint:**For an adiabatic process there is no heat exchange and to calculate heat from isobaric and isochoric(isovolumetric) process use molar heat capacity at constant pressure and molar heat capacity at constant volume and from the equation of adiabatic expansion you can find the ratio of molar heat capacity at constant pressure and molar heat capacity at constant volume.

**Complete step by step answer:**

Let look at the adiabatic equation, we have ${P^3}{V^5} = {\rm{constant}}$ which can be also written as $P{V^{\dfrac{5}{3}}} = {\rm{constant}}$ therefore, $\gamma = \dfrac{5}{3}$ from the value of $\gamma$ we can find molar heat capacity at constant pressure and molar heat capacity at constant volume i.e; ${C_p} \& {C_V}$.

So, $\gamma = \dfrac{{{C_P}}}{{{C_V}}} = \dfrac{5}{3}$

$ \Rightarrow {C_P} = \dfrac{5}{2}R \& {C_V} = \dfrac{3}{2}R$

Now will find heat supplied in both cases

For case(i) isobaric process

Heat supplied ${H_1} = n{C_P}\Delta T$

$ \Rightarrow {H_1} = n\dfrac{5}{2}R({T_f} - {T_i})\\

\Rightarrow {H_1} = \dfrac{5}{2}(nR{T_f} - nR{T_i})\\

\Rightarrow {H_1} = \dfrac{5}{2}({P_i}{V_f} - {P_i}{V_i})$

Putting value of pressure and volume we have value of heat supplied as

${H_1} = \dfrac{5}{2} \times {10^5}(8 - 1){10^{ - 3}}\\

\Rightarrow{H_1}= 1750\,joule$

For case(ii) isochoric(isovolumetric)

Heat supplied ${H_2} = n{C_V}\Delta T$

$ \Rightarrow {H_2} = n\dfrac{3}{2}R({T_f} - {T_i})\\

\Rightarrow {H_2} = \dfrac{5}{2}(nR{T_f} - nR{T_i})\\

\Rightarrow {H_2} = \dfrac{5}{2}({P_f}{V_f} - {P_i}{V_f})$

Putting values of pressure and temperature we have,

${H_2} = \dfrac{3}{2} \times 8 \times {10^{ - 3}} \times (\dfrac{1}{{32}} - 1){10^5}\\

\Rightarrow{H_2}= - 1162 joule$

Total heat supplied will be sum of these two hearts

$H = {H_1} + {H_2} \\

\Rightarrow H= 1750 + ( - 1162) \\

\therefore H= 588\,joule$

**Hence, the correct answer is Option C.**

**Note:**Here expression of adiabatic expansion is given to help you find molar heat capacity at constant pressure and molar heat capacity at constant volume because you don’t know the which type of gas is it although you know it once you found the molar heat capacity at constant pressure and molar heat capacity at constant volume. Once you know these two heat capacities simply put the formula to get the final answer.

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