Question

If a gas is taken from $A$ to $C$ through $B$, then heat absorbed by the gas is $8\text{J}$. Heat absorbed by the gas in taking it directly from $A$ to $C$ is

• A. $+8\text{J}$
• B. $+9\text{J}$
• C. $+5\text{J}$
• D. $+7\text{J}$

Answer 1

The correct option is B $+9\text{J}$

When the gas is taken through process $ABC$,
Work done is during the process is given by
$W=W_{AB}+W_{BC}$
But, $W_{BC}=0$ [isochoric process]
Thus,
$W=P\Delta V=10\times [400-200]\times [{10}^3\times {10}^{-6}]=2\text{J}$
According to first law of thermodynamics,
Heat absorbed $\Delta Q=\Delta U+W$
From the data given in the question,
$\Rightarrow 8=\Delta U+W$
$\Rightarrow 8=\Delta U+2$
$\Rightarrow \Delta U=6\text{J}$
Here initial state is $A$ and final state is $C$
Thus $\Delta U=U_C-U_A=+6\text{J}.....\left(1\right)$

Now, when directly taken from $A$ to $C$, by applying first law of thermodynamics, we get
$\Delta Q^{\prime }=W^{\prime }+\Delta U$
Using$\left(1\right)$,
$\Delta Q^{\prime }=W^{\prime }+6......\left(2\right)$
Here $W^{\prime }$ is area under process $AC$.
So, $W^{\prime }=\left(\frac{1}{2}\right[(20+10)\times {10}^3\left]\right[(400-200)\times {10}^{-6}\left]\right)$
$\Rightarrow W^{\prime }=+3\text{J}$
Substituting this in $\left(2\right)$ we get,
$\Delta Q^{\prime }=6+3=+9\text{J}$
Thus, option (b) is the correct answer.