A thermodynamic system undergoes a cyclic pro | vedclass

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

Question

(B) The work done in a cyclic process is equal to the area enclosed by the $P-V$ loop.Since the cycle $PQRSP$ is traversed in an anti-clockwise direct

Vedclass · Accepted Answer

$-20$

Vedclass · Answer

(B) The work done in a cyclic process is equal to the area enclosed by the $P-V$ loop.Since the cycle $PQRSP$ is traversed in an anti-clockwise direction, the work done by the system is negative.Area $= 	ext{length} 	imes 	ext{width} = (V_Q - V_P) 	imes (P_S - P_P)$$V_Q - V_P = (300 - 100) 	ext{ cc} = 200 	ext{ cc} = 200 	imes 10^{-6} 	ext{ m}^3$$P_S - P_P = (200 - 100) 	ext{ kPa} = 100 	ext{ kPa} = 100 	imes 10^3 	ext{ Pa}$Work done $W = - (	ext{Area}) = - (200 	imes 10^{-6} 	ext{ m}^3) 	imes (100 	imes 10^3 	ext{ Pa})$$W = - (200 	imes 100 	imes 10^{-3}) 	ext{ J} = -20 	ext{ J}$.

$A$ thermodynamic system undergoes a cyclic process $PQRSP$. The total work done by the system is..... $J$.

Similar Questions

An ideal gas of mass $m$ in a state $A$ goes to another state $B$ via three different processes as shown in the figure. If $Q_{1}, Q_{2}$,and $Q_{3}$ denote the heat absorbed by the gas along the three paths,then

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

$A$ thermodynamic system is taken through the cycle $PQRSP$ process as shown in the figure. The net work done by the system is .... $J$. (Note: $1 \text{ kp} = 10^3 \text{ Pa}$,$1 \text{ cc} = 10^{-6} \text{ m}^3$)

$A$ system is taken through a cyclic process represented by a circle as shown. The heat absorbed by the system is

The process $CD$ is shown in the diagram. As the system is taken from $C$ to $D$,what happens to the temperature of the system?

Vedclass Test Series

Exam Paper Generator

Online Exam Module