Question

A quantity of air is taken from state a to state b along a path that is a straight line in the P-V diagram. In the previous question If Va = 0.0700 $m^3$, Vb = 0.1100 $m^3$, Pa = 1.00 $\times$ ${10}^5$ Pa, and Pb = 1.40 $\times$${10}^5$ Pa, what is the work W done by the gas in this process? Assume that the gas may be treated as ideal.

• A. 12000 J
• B. 4800 J
• C. 1200 J
• D. 5400 J

Answer 1

The correct option is B

4800 J

The area under the P-V graph will be the amount of work done. In this figure -

the work done = area of rectangle (acde) + area of $\Delta \left(abc\right)$

$=(V_b-V_a\times P_a+\left(\frac{1}{2}\right)(V_b-V_a)\times (P_b-P_a).$

Substituting with known values we get -

$workdone=W=\left[\right(0.11-0.07)\times {10}^5+\left(\frac{1}{2}\right)(0.11-0.07)\times 0.4\times {10}^5]J$

$=(4\times {10}^3+800)J=4800J.$