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 $\mathrm{B}$ to $\mathrm{C}$ 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) $33800 \mathrm{~J}$
(2) $2200 \mathrm{~J}$
(3) $600 \mathrm{~J}$
(4) $800 \mathrm{~J}$
Show Answer
Answer: (4)
Solution:
Work done $\mathrm{AB}=\frac{1}{2}(8000+6000)$ Dyne $/ \mathrm{cm}^{2} \times$
$4 \mathrm{~m}^{3}=\left(6000\right.$ Dyne $\left./ \mathrm{cm}^{2}\right) \times 4 \mathrm{~m}^{3}$
Work done $\mathrm{BC}=-\left(4000 \mathrm{Dyne} / \mathrm{cm}^{2}\right) \times 4 \mathrm{~m}^{3}$
Total work done $=2000$ Dyne $/ \mathrm{cm}^{2} \times 4 \mathrm{~m}^{3}$
$=2 \times 10^{3} \times \frac{1}{10^{5}} \frac{\mathrm{N}}{\mathrm{cm}^{2}} \times 4 \mathrm{~m}^{3}$
$=2 \times 10^{-2} \times \frac{\mathrm{N}}{10^{-4} \mathrm{~m}^{2}} \times 4 \mathrm{~m}^{3}$
$=2 \times 10^{2} \times 4 \mathrm{Nm}=800 \mathrm{~J}$
