The pressure and volume of a gas are changed along a path ABCA in the figure. The vertical divisions on the graph represent 2.5 105 Pa, and the horizontal divisions represent 1.0 10-3 m3. Determine the work done (including algebraic sign) in each segment of the path.
a. A to b
b. B to C
c. c to A
itis in a shape of a right-angled triangle turn upside down
i solved it. Thanks. To find work W=PV ie Multiply the P by V given. count the squares beneathe the line and multiply by the work. If it's a vertical line, it has no squares beneathe, thus, it's 0.
To determine the work done along each segment of the path, we can use the formula W = P * ∆V, where W is the work done, P is the pressure, and ∆V is the change in volume.
a. A to B: This segment is a horizontal line, meaning there is no change in volume (∆V = 0). Therefore, the work done is also zero: W = P * ∆V = P * 0 = 0.
b. B to C: This segment is represented by the vertical line. To determine the change in volume, we need to count the number of squares beneath the line. In this case, it looks like there are 4 squares. Each horizontal division represents 1.0 * 10^-3 m^3, so the change in volume (∆V) is 4 * 1.0 * 10^-3 m^3 = 4 * 10^-3 m^3. Multiply this by the pressure (P) to find the work done: W = P * ∆V = P * (4 * 10^-3 m^3).
c. C to A: This segment is represented by the hypotenuse of the right-angled triangle. To find the change in volume, we need to determine the length of this hypotenuse. Count the number of vertical and horizontal divisions, which in this case, seems to be 6 vertical squares and 8 horizontal squares. Each vertical division represents 2.5 * 10^5 Pa, and each horizontal division represents 1.0 * 10^-3 m^3. Therefore, the change in volume (∆V) is 6 * 2.5 * 10^5 Pa * 8 * 1.0 * 10^-3 m^3. Multiply this by the pressure (P) to find the work done: W = P * ∆V = P * (6 * 2.5 * 10^5 Pa * 8 * 10^-3 m^3).
Remember to include the appropriate algebraic sign (+ or -) based on the direction of the process (expansion or compression).