Chapter 1: Q. 1.33 (page 1)

An ideal gas is made to undergo the cyclic process shown in the given figure. For each of the steps A, B, and C, determine whether each of the following is positive, negative, or zero: (a) the work done on the gas; (b) the change in the energy content of the gas; (c) the heat added to the gas.
Then determine the sign of each of these three quantities for the whole cycle. What does this process accomplish?

Short Answer

Expert verified

a. Total work done on the gas is $W_{iotal} = \frac{1}{2} (P_{2} - P_{1}) (V_{2} - V_{1}) > 0 .$

b. There is no net change in thermal energy.

c. Total heat added is $- W$ .

Step by step solution

Given information

An ideal gas is made to undergo the cyclic process shown in the figure

Calculation

$Work done for side A is W_{A} = - \int_{V_{d}}^{V_{f}} P_{1} d V W_{A} = - P_{1} (V_{2} - V_{1}) That is actually negative quantity i.e. works done for part A is negative because the pressure is constant and V_{2} > V_{1} . For part B, volume is constant throughout, so work done is W_{B} = 0 \dots \dots \dots (3) For side C, the equation for pressure is P - P_{1} = (\frac{p_{2} - p_{1}}{V_{2} - V_{1}}) (V - V_{1}) \dots \dots . (3) Here (\frac{P_{2} - P_{1}}{V_{2} - V_{1}}) is the slope. From equation (3), expression for P can be obtained as : P = (\frac{P_{2} - P_{1}}{V_{2} - V_{1}}) (V - V_{1}) + P_{1} Substitute (\frac{P_{2} - P_{1}}{V_{2} - V_{1}}) (V - V_{1}) + P_{1} for P in equation (1) to obtained work done for side C. W_{C} = - \int_{V_{2}}^{V_{1}} {[(\frac{P_{2} - P_{1}}{V_{2} - V_{1}}) (V - V_{1}) + P]}_{1} d V \dots \dots (5) Exchanging the integration limit of equation (5) we get W_{C} = \int_{V_{1}}^{V_{2}} {[(\frac{P_{2} - P_{1}}{V_{2} - V_{1}}) (V - V_{1}) + P]}_{1} d V$

$W_{C} = \int_{V_{1}}^{V_{2}} {[(\frac{P_{2} - P_{1}}{V_{2} - V_{1}}) (V - V_{1}) + P_{1}]}_{1} d V W_{C} = \int_{V_{1}}^{V_{2}} {[(\frac{P_{2} - P_{1}}{V_{2} - V_{1}}) V - (\frac{P_{2} - P_{1}}{V_{2} - V_{1}}) V_{1} + P_{1}]}_{1} d V W_{C} = {[(\frac{p_{2} - p_{1}}{V_{2} - V_{1}}) \frac{V^{2}}{2} - (\frac{p_{2} - p_{1}}{V_{2} - V_{1}}) V_{1} V + P_{1} V]}_{V_{1}}^{V_{2}} W_{C} = [(\frac{p_{2} - p_{1}}{V_{2} - V_{1}}) \frac{V_{2} {}^{2}- {V_{1}}^{2}}{2} - V_{1} (\frac{p_{2} - p_{1}}{V_{2} - V_{1}}) (V_{2} - V_{1}) + P_{1} (V_{2} - V_{1})] \dots \dots . . (6) But V_{2} {}^{2}- V_{1} {}^{2}= (V_{2} - V_{1}) (V_{2} + V_{1})$

$Equation (6) becomes W_{C} = [\frac{(P_{2} - P_{1}) V_{2} + V_{1})}{2} - V_{1} (\frac{P_{2} - P_{1}}{V_{2} - V_{1}}) (V_{2} - V_{1}) + P_{1} (V_{2} - V_{1})] W_{C} = [\frac{(P_{2} - p_{1}) (V_{2} + V_{1})}{2} - V_{1} (\frac{p_{2} - P_{1}}{V_{2} - V_{1}}) (V_{2} - V_{1}) + P_{1} (V_{2} - V_{1})] W_{C} = [(P_{2} - P_{1}) (\frac{V_{2}}{2} + \frac{V_{1}}{2} - V_{1}) + P_{1} (V_{2} - V_{1})] W_{C} = [(P_{2} - P_{1}) (\frac{V_{2}}{2} - \frac{V_{1}}{2}) + P_{1} (V_{2} - V_{1})] W c = [\frac{1}{2} (P_{2} - P_{1}) (V_{2} - V_{1}) + P_{1} (V_{2} - V_{1})] \dots \dots . . (7)$

$Since P_{2} > P_{1} and V_{2} > V_{1}, so W_{C} > 0 Total work done on the gas is: W = W_{A} + W_{B} + W_{C} \dots \dots . . (8) Substitute - P_{1} (V_{2} - V_{1}) for W_{A}, 0 for W_{B} and [\frac{1}{2} (P_{2} - P_{1}) (V_{2} - V_{1}) + P_{1} (V_{2} - V_{1})] for W_{C} in equation (8) W = \frac{1}{2} (P_{2} - P_{1}) (V_{2} - V_{1}) + P_{1} (V_{2} - V_{1}) - P_{1} (V_{2} - V_{1}) W_{total} = \frac{1}{2} (P_{2} - P_{1}) (V_{2} - V_{1}) > 0$

Hence, total work done on the gas is $W_{iotal} = \frac{1}{2} (P_{2} - P_{1}) (V_{2} - V_{1}) > 0 .$

Continuation of calculation

Change in Thermal energy along the side A is :-

$Δ U_{A} = \frac{3}{2} P Δ V Δ U_{A} = \frac{3}{2} P_{1} (V_{2} - V_{1}) Δ U_{A} = \frac{3}{2} P_{1} (V_{2} - V_{1}) > 0$

Since pressure is constant and volume increases along the side A as shown in figure above.

Along the side B, volume is constant, and pressure increases so change in thermal energy for side B is therefore :

$Δ U_{B} = \frac{3}{2} V Δ P Δ U_{B} = \frac{3}{2} V_{2} (P_{2} - P_{1}) Δ U_{B} = \frac{3}{2} V_{2} (P_{2} - P_{1}) > 0$

Along C, both pressure and volume decrease so change in thermal energy for side C is

$Δ U_{C} = \frac{3}{2} Δ (P V) Δ U_{C} = \frac{3}{2} Δ (P_{1} V_{1} - P_{2} V_{2}) Δ U_{C} = - \frac{3}{2} (P_{2} V_{2} - P_{1} V_{1}) < 0$

Total thermal energy is $U = U_{Λ} + U_{B} + U_{C}$

Substitute $\frac{3}{2} P_{1} (V_{2} - V_{1}) for Δ U_{A}, \frac{3}{2} V_{2} (P_{2} - P_{1}) for Δ U_{B} and - \frac{3}{2} (P_{2} V_{2} - P_{1} V_{1}) for Δ U_{C} in equation (5)$

$U_{tokal} = \frac{3}{2} [P_{1} (V_{2} - V_{1}) + V 2 (P_{2} - P_{1}) - (P_{2} V_{2} - P_{1} V_{1})] U_{total} = 0$

Hence, the net change in thermal energy is $U_{toual} = 0 .$

Continuation of calculation

$S u b s t i t u t e 0 f o r W_{B} a n d \frac{3}{2} V_{2} (P_{2} - P_{1}) f o r Δ U_{B} i n e q u a t i o n (1) f o r s i d e B Q_{B} = \frac{3}{2} V_{2} (P_{2} - P_{1}) + 0 Q_{B} = \frac{3}{2} V_{2} (P_{2} - P_{1}) S i n c e t o t a l w o r k d o n e f o r s i d e C i s f o u n d f r o m a b o v e c a l c u l a t i o n i . e . W_{C} = [\frac{1}{2} (P_{2} - P_{1}) (V_{2} - V_{1}) + P_{1} (V_{2} - V_{1})] a n d n e t c h a n g e i n t h e r m a l e n e r g y i s f o u n d a s Δ U_{C} = - \frac{3}{2} (P_{2} V_{2} - P_{1} V_{1})$

$S u b s t i t u t e [\frac{1}{2} (P_{2} - P_{1}) (V_{2} - V_{1}) + P_{1} (V_{2} - V_{1})] f o r W_{C} a n d - \frac{3}{2} (P_{2} V_{2} - P_{1} V_{1}) f o r Δ U_{C} i n e q u a t i o n (1) f o r s i d e C Q_{C} = - [\frac{1}{2} (P_{2} - P_{1}) (V_{2} - V_{1}) + P_{1} (V_{2} - V_{1})] - \frac{3}{2} (P_{2} V_{2} - P_{1} V_{1}) T o t a l h e a d a d d e d i s c a l c u l a t e d a s Q_{totat} = Q_{Λ} + Q_{B} + Q_{C} \dots \dots . . . . (5) S u b s t i t u t e \frac{5}{2} P_{1} (V_{2} - V_{1}) f o r Q_{A}, \frac{3}{2} V_{2} (P_{2} - P_{1}) f o r Q_{B} a n d - [\frac{1}{2} (P_{2} - P_{1}) (V_{2} - V_{1}) + P_{1} (V_{2} - V_{1})] - \frac{3}{2} (P_{2} V_{2} - P_{1} V_{1}) f o r Q_{C} i n e q u a t i o n (5) Q = \frac{5}{2} P_{1} (V_{2} - V_{1}) + \frac{3}{2} V_{2} (P_{2} - P_{1}) - [\frac{1}{2} (P_{2} - P_{1}) (V_{2} - V_{1}) + P_{1} (V_{2} - V_{1})] - \frac{3}{2} (P_{2} V_{2} - P_{1} V_{1}) Q_{total} = - \frac{1}{2} (P_{2} - P_{1}) (V_{2} - V_{1}) = - W$

Final answer

a. Total work done on the gas is $W_{iotal} = \frac{1}{2} (P_{2} - P_{1}) (V_{2} - V_{1}) > 0 .$

b. There is no net change in thermal energy.

c. Total heat added is $- W .$

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