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 $/ cm^{2} \times 4 m^{3}$
$=2 \times 10^{3} \times \frac{1}{10^{5}} \frac{N}{cm^{2}} \times 4 m^{3}$
$=2 \times 10^{-2} \times \frac{N}{10^{-4} m^{2}} \times 4 m^{3}$
$=2 \times 10^{2} \times 4 Nm=800 J$
