Question

5. Find the quasi-static (reversible) work done and the net heat absorbed by the system40.00 in each of the following three processes, all of which take the system from state A to state B Please refer to the diagram on the right. The blue curve and the equation that refers to it are for an adiabatic process. a is a constant. 2.63. a) The system is expanded from its original to its final volume heat being added to maintain the pressure constant. The volume i then kept constant, and heat is extracted to reduce the pressure to 2.63 atm. 1.00 8.00 V(L) b) The volume is increased, and heat is supplied to cause the pressure to decrease linearly with the volume c) The two steps of process (a) are performed in the opposite order (but in the same direction)

Answer 1

(a)

$dW=PdV$

The process is from A to B` and from B` to B.

From A to B` , the process is at constant pressure, so work done,

$W_1=P\int dV=40\int_1^8 dV$

W = 40 x 7 = 280 J

From B` to B, the process is at constant volume, hence dV=0 and so work done =

$W_2=\int PdV=0$

Total Work = W₁ + W₂ = 280 J

(b)

The process is through the straight blue line from to A to B.

Slope of line,

$m=\frac{2.63-40}{8-1}=-5.33$

Hence, we can write the equation as,

(P - 2.63) = - 5.33(V - 8)

P = -5.33V + 45.33

Work done in going from A to B,

$W=\int PdV= \int_1^8(-5.33V+45.33)dV$

Integrating, we get,

W = 149.41 J

(c)

$dW=PdV$

The process is from A to A` and from A` to B.

From A to A` , the process is at constant volume, hence dV=0 and so work done =
$W_1=\int PdV=0$
From A` to B, the process is at constant pressure,

$W_1=P\int dV=2.63\int_1^8 dV$

W = 2.63 x 7 = 18.41 J

Total Work = W₁ + W₂ = 18.41 J