Question 254: Thermodynamics & KTG

Question

A thermodynamic system is taken from an original state $$\mathrm{A}$$ 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 :

Accepted Answer

Work done $$\mathrm{AB}=\frac{1}{2}(8000+4000)$$ Dyne$$/\mathrm{cm}^2 	imes 4 \mathrm{~m}^3=\left(6000ight.$$ Dyne$$\left./\mathrm{cm}^2ight) 	imes 4 \mathrm{~m}^3$$
Work done $$\mathrm{BC}=-\left(4000ight.$$ Dyne$$\left./\mathrm{cm}^2ight) 	imes 4 \mathrm{~m}^3$$
Total work done $$=2000$$ Dyne$$/\mathrm{cm}^2 	imes 4 \mathrm{~m}^3$$
$$\begin{aligned}
=2 	imes 10^3 & 	imes \frac{1}{10^5} \frac{\mathrm{N}}{\mathrm{cm}^2} 	imes 4 \mathrm{~m}^3 \
& =2 	imes 10^{-2} 	imes \frac{\mathrm{N}}{10^{-4} \mathrm{~m}^2} 	imes 4 \mathrm{~m}^3 \
& =2 	imes 10^2 	imes 4 \mathrm{Nm}=800 \mathrm{~J}
\end{aligned}$$

Answer

800 J

Answer

2200 J

Answer

33800 J

Answer

1200 J

Thermodynamics & KTG

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