### Thermodynamics Question 5

#### Question 5 - 2024 (29 Jan Shift 1)

A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as shown in the figure. It’s volume is then reduced to the original value from $B$ to $C$ by an isobaric process. The total work done by the gas from $A$ to $B$ and $B$ to $C$ would be :

[We changed options. In official NTA paper no option was correct.]

(1) $33800 J$

(2) $2200 J$

(3) $600 J$

(4) $800 J$

## Show Answer

#### Answer: (4)

#### Solution:

Work done $AB=\frac{1}{2}(8000+6000)$ Dyne $/ cm^{2} \times$

$4 m^{3}=\left(6000\right.$ Dyne $\left./ cm^{2}\right) \times 4 m^{3}$

Work done $BC=-\left(4000 Dyne / cm^{2}\right) \times 4 m^{3}$

Total work done $=2000$ Dyne $/ cm^{2} \times 4 m^{3}$

$=2 \times 10^{3} \times \frac{1}{10^{5}} \frac{N}{cm^{2}} \times 4 m^{3}$

$=2 \times 10^{-2} \times \frac{N}{10^{-4} m^{2}} \times 4 m^{3}$

$=2 \times 10^{2} \times 4 Nm=800 J$