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A system is taken from state a to state $c$ by two paths $a d c$ and $a b c$ as shown in the figure. The internal energy a is $U_2=10$ J. Along the path adc the amount of heat heat absorbed $\delta Q_1=50 \mathrm{~J}$ and the work obtained $\delta W_1=20 \mathrm{~J}$ whereas along the path $a b c$ the heat absorbed $\delta Q_2=36 \mathrm{~J}$. The amount of work along the path $a b c$ is:
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The correct answer is:
$6 \mathrm{~J}$
According to $1^{\text {st }} 1$ aw of thermodynamics
$\delta Q=\delta V+\delta W$
Along the path adc
$\delta U_1=\delta Q_1-\delta W_1=50-20=30 \mathrm{~J}$
then path abc
Since change in Internal energy in path in dependence
$\begin{aligned}
\text {So, } \delta U_1 & =\delta U_2 \\
30 \mathrm{~J} & =36 \mathrm{~J}-\delta W_2 \\
\Rightarrow \delta W_2 & =36 \mathrm{~J}-30 \mathrm{~J} \\
& =6 \mathrm{~J}
\end{aligned}$
$\delta Q=\delta V+\delta W$
Along the path adc
$\delta U_1=\delta Q_1-\delta W_1=50-20=30 \mathrm{~J}$
then path abc
Since change in Internal energy in path in dependence
$\begin{aligned}
\text {So, } \delta U_1 & =\delta U_2 \\
30 \mathrm{~J} & =36 \mathrm{~J}-\delta W_2 \\
\Rightarrow \delta W_2 & =36 \mathrm{~J}-30 \mathrm{~J} \\
& =6 \mathrm{~J}
\end{aligned}$
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