In the following p-V diagram, the equation of | vedclass

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

Question

(D) The work done in a closed $p-V$ cycle is equal to the area enclosed by the path. Since the cycle is traversed in a counter-clockwise direction, th

Vedclass · Accepted Answer

$-\frac{1}{3a}$

Vedclass · Answer

(D) The work done in a closed $p-V$ cycle is equal to the area enclosed by the path. Since the cycle is traversed in a counter-clockwise direction, the work done by the gas is negative.The area under the curve $(V-2)^2 = 4aP$ from $V=1$ to $V=3$ is given by $\int_{1}^{3} P \, dV$.From the equation, $P = \frac{(V-2)^2}{4a}$.Area under the curve $= \int_{1}^{3} \frac{(V-2)^2}{4a} \, dV = \frac{1}{4a} \left[ \frac{(V-2)^3}{3} ight]_{1}^{3} = \frac{1}{12a} [(3-2)^3 - (1-2)^3] = \frac{1}{12a} [1 - (-1)] = \frac{2}{12a} = \frac{1}{6a}$.At $V=1$ or $V=3$, the pressure $P_0$ is given by $(1-2)^2 = 4aP_0$, so $P_0 = \frac{1}{4a}$.The area of the rectangle formed by the top horizontal line (at $P_0$) and the $V$-axis from $V=1$ to $V=3$ is $P_0 	imes (3-1) = \frac{1}{4a} 	imes 2 = \frac{1}{2a}$.The area enclosed by the cycle is the area of the rectangle minus the area under the curve:Area $= \frac{1}{2a} - \frac{1}{6a} = \frac{3-1}{6a} = \frac{2}{6a} = \frac{1}{3a}$.Since the cycle is counter-clockwise, the work done $W = -\frac{1}{3a}$.

In the following $p-V$ diagram, the equation of state along the curved path is given by $(V-2)^2 = 4aP$, where $a$ is a constant. The total work done in the closed path is:

Similar Questions

The $P-V$ diagram of an ideal gas is as shown in the figure. The work done by the gas in the process $ABCD$ is:

$A$ thermodynamic system undergoes a cyclic process $ABCDA$ as shown in the figure. The work done by the system in the cycle is

One mole of an ideal gas is taken from $a$ to $b$ along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is $w_s$ and that along the dotted line path is $w_d$,then the integer closest to the ratio $w_d / w_s$ is

$A$ system goes from $A$ to $B$ via two processes $I$ and $II$ as shown in the figure. If $\Delta U_1$ and $\Delta U_2$ are the changes in internal energies in the processes $I$ and $II$ respectively,then:

An ideal gas is taken through a process $ABCA$ as shown in the figure. If the net heat supplied to the gas in the cycle is $5 \ J$,then the work done by the gas in the process from $C$ to $A$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module