Question

A figure show a closed cycle for a gas. The change in internal energy along the path $ca$ is $-160\text{J}$. The energy transferred to the gas as heat is $200\text{J}$ along path $ab$ and $40\text{J}$ along path $bc$. What is the ratio of work done by the gas along the path $abc$ and $ab$?

• A. 1 : 1
• B. 2 : 1
• C. 1 : 2
• D. 3 : 2

Answer 1

The correct option is A 1 : 1

Given that
$\Delta U_{ca}=U_a-U_c=-160\text{J}$
$\Rightarrow \Delta U_{ac}=U_c-U_a=160\text{J}$
$Q_{bc}=40\text{J}$
$Q_{ab}=200\text{J}$

Since internal energy only depends on initial and final states,
$\Delta U_{ac}=\Delta U_{ab}+\Delta U_{bc}$
$\Rightarrow 160=\Delta U_{ab}+\Delta U_{bc}\dots \left(1\right)$

Using first law of thermodynamics,
$\Delta U_{ab}=Q_{ab}-W_{ab};\Delta U_{bc}=Q_{bc}-W_{bc}$
From (1),
$160=Q_{bc}-W_{bc}+Q_{ab}-W_{ab}$
$\Rightarrow 160=40-0+200-W_{ab}$
$\Rightarrow W_{ab}=80\text{J}$
[Since there is no change in volume in path $bc$ and work done is zero]

So, $W_{ab}=W_{abc}=80\text{J}$
$\therefore \frac{W_{ab}}{W_{bc}}=\frac{80\text{J}}{80\text{J}}=1$
Hence, option $\left(a\right)$ is correct.