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Isobaric expansion of gas.
As P is constant,  V is proportional to the temperature T.

\frac{V_2}{V_1}=\frac{T_2}{T_1}\\\\V_2=\frac{V1T_2}{T_1}\\\\W= \int\limits^{V_2}_{V_1} {P} \, dV =P \times (V_2-V_1)\\\\W=PV_1\ \frac{T_2-T_1}{T_1}=nRT_1*\frac{T_2-T_1}{T_1}=nR(T_2-T_1)\\\Delta U=n C_v(T_2-T_1)=\frac{nC_p}{\gamma}(T_2-T_1)\\\\\Delta Q=Heat\ supplied=nC_p(T_2-T_1)=100J\\\\\Delta U=\frac{\Delta Q}{\gamma}=100/(5/3)=60Joules\\\\Work\ done=100-60=40\ Joules\\\\\Delta Q=W+\Delta U

Also, we find that  C_p / C_v = 5/3
C_p - C_v = R
C_v [5/3 - 1] = R  =>  C_v = 3/2 R
        => C_p = 5/2 R

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