Question

Example An ideal gas is expanded from 1 m to 4 m in a 3-step process Step I: Expansion from I m to 2 m ata pressure of 6 M Pa. Step 2: Expansion from 2 m to 3m as the Step 3: Expansion from 3 m to 4 mat a pressure a) What is the total work done on the gas during the expansion? ) If the process were reversed (ie. 3-step contraction) pressure is decreased uniformly from 6 M Pa to 2 M Pa of 2 M Pa. b. what would the total work be?

Answer 1

Work done on the step -1 is (It is basically a constant pressure process)

$W_1 = P. dV$

$W_1 = (6 \times 10^6 Pa} (2 m^3 - 1 m^3)$

$W_1 = 6 MJ$

Work done on the step -2 is

$W_2 = ( 6 MPa - 2 MPa) (3 m^3 - 2 m^3)$

$W_2 = 4 MJ$

Similarly, Work done in the third process

$W_2 = 2 MPa ( 4 m^3 - 3 m^3)$

$W_3 = 2 MJ$

So, the total work done is

W = (6 + 4 + 2 ) MJ

W = 12 MJ