Question

A sample of an ideal gas is taken through the cyclic process \(abca\) as shown in the figure. The change in the internal energy of the gas along the path ca is \(-180~\text J,\) The gas absorbs \(250~\text J\) of heat along the path \(ab\) and \(60~\text J\) along the path \(bc\). The work down by the gas along the path \(abc\) is :

\( V \longrightarrow \)

• A. $120\text{J}$
• B. $130\text{J}$
• C. $140\text{J}$
• D. $100\text{J}$

Answer 1

The correct option is B $130\text{J}$

As the process $abca$ is cyclic,

$\Delta U_{abca}=0$

$\Rightarrow \Delta U_{abc}+\Delta U_{ca}=0$

$\Rightarrow \Delta U_{abc}=-(-180)=180\text{J}$

$\text{Also,}{Q}_{abc}=Q_{ab}+Q_{bc}$

$\Rightarrow Q_{abc}=250+60=310\text{J}$

Now, using first law of thermodynamics,

$Q_{abc}=\Delta U_{abc}+W_{abc}$

$\Rightarrow 310=180+W_{abc}$

$\Rightarrow W_{abc}=130\text{J}$

Hence, option $\left(B\right)$ is correct.