A thermodynamic system is taken from an origi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The work done in a $P-V$ diagram is equal to the area under the curve.For the process $A \rightarrow B$,the work done is the area of the trapezoid

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(C) The work done in a $P-V$ diagram is equal to the area under the curve.For the process $A ightarrow B$,the work done is the area of the trapezoid under the line $AB$:$W_{AB} = 	ext{Area of trapezoid} = \frac{1}{2} 	imes (P_A + P_B) 	imes (V_B - V_A)$$W_{AB} = \frac{1}{2} 	imes (8000 + 4000) 	ext{ dyne/cm}^2 	imes (7 - 3) 	ext{ m}^3 = 6000 	ext{ dyne/cm}^2 	imes 4 	ext{ m}^3 = 24000 	ext{ dyne} \cdot 	ext{m}^3/	ext{cm}^2$.Converting units: $1 	ext{ dyne/cm}^2 = 0.1 	ext{ N/m}^2$. So,$W_{AB} = 24000 	imes 0.1 	ext{ J} = 2400 	ext{ J}$.For the process $B ightarrow C$,the work done is the area under the line $BC$ (isobaric compression):$W_{BC} = P_B 	imes (V_C - V_B) = 4000 	ext{ dyne/cm}^2 	imes (3 - 7) 	ext{ m}^3 = 4000 	imes (-4) 	ext{ dyne} \cdot 	ext{m}^3/	ext{cm}^2 = -16000 	ext{ dyne} \cdot 	ext{m}^3/	ext{cm}^2$.Converting units: $W_{BC} = -16000 	imes 0.1 	ext{ J} = -1600 	ext{ J}$.Total work done $W = W_{AB} + W_{BC} = 2400 	ext{ J} - 1600 	ext{ J} = 800 	ext{ J}$.

$A$ thermodynamic system is taken from an original state $A$ to an intermediate state $B$ by a linear process as shown in the figure. Its 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 $C$ is: (in $J$)

Similar Questions

An ideal gas follows the path shown in the figure. The net work done in the whole cycle is

$A$ monoatomic ideal gas goes through a cyclic process as shown in the figure. The efficiency of this process is (in $\%$)

Find the work done in the cyclic process $ABCA$ shown in the figure.

One mole of an ideal gas is taken from $A$ to $B$,from $B$ to $C$,and then back to $A$. The variation of its volume with temperature for that change is as shown. Its pressure at $A$ is $P_{0}$ and volume is $V_{0}$. Then,the internal energy:

The heat energy absorbed by a system in going through a cyclic process shown in the figure is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module