(IV) A fancy box holds a fixed amount of ideal diatomic gas. The gas passes thro
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(IV) A fancy box holds a fixed amount of ideal diatomic gas. The gas passes through the sequence of states shown in the figure. (Vi , F)-(1/5.10000) (V, P) (1/2, 4000) (G. B)-(1/5, 400) PV=nRT Pi (a) Compute the temperature of the gas in the three "corner" states. (b) Compute the internal energy of the gas in the three "corner" states (c) In the transition from 1 2 ie, (.P) (V.Ps), alonug the diagoual line in the T(2)- T(3) U(2) U(3) (6) An amount of work was done CHOOSE ONE:] BY ON the gas. (ii) From the temperatures and internal energies arrived at in (a) and (b), and the low the gas. of work, we can infer that heat Bowed [CHOOSE ONE] out of (ii) In the transition from 12, the entropy of the gas in the box [CHOOSE ONE:1 INCREASED REMA?NED CONSTANT (b) Consider the transition from 2 ? 3 [leftward along the horizontal line in the figure. (0) An amount of work was done CHOOSE ONE: BY ON the gas. (ü) From the temperatures and internal energies arrived at in (a) and (b), and the fow the gas. of work, we can infer that heat flowed (CuooSE ONE:] into out of (ii) In the transition from 2-+3, the entropy of the gas in the box [CHOOSE ONE] INCREASED REMAINED CONSTANT (0) Determine the work done by the gas during this transition. (G) From the temperatures and internal energies arrived at in (a) and (b), and the Bow (c) Consider the transition from 3-+1 lupward along the vertical line in the figure). into the gas. of work, we can infer that heat flowed [CHoosE ONE: (ii) In the transition from 3+1, the entropy of the gas in the box [CHOOSE ONE] INCREASED REMAINED CONSTANT DECREASED Page 3 of 4Explanation / Answer
iv) FOR THE GIVEN DIATOMIC GAS
(V1, P1) = (1/5, 10000)
(V2, P2) = (1/2, 4000)
(V3, P3) = (1/5, 4000)
a. using PV = nRT
T1 = P1V1/nR
nR = 8 J / K
hence
T1 = P1V1/8 = 250 K
similiarly
T2 = P2V2/8 = 250 K
similiarly
T3 = P3V3/8 = 100 K
b. U(1) = 5 nRT/2 = 20T = 20T1 = 5000J
U(2) = 5000 J
U(3) = 2000 J
c. for process 1 -> 2
volume of the gas increases
i. hence, amoun tof work was done by the gas
ii. now, work done by gas > 0
U(2) - U(1) = 0
hence heat flow is into the gas
beasuee heat flow into the gas H = dU + w
W > 0 hence H > 0
iii. in transition 1 -> 2
heat flows into the gas, so entropy of the gas increases
b. form 2 -> 3
i. volume of the gas decreases
so work is done on the gas
ii. dU = U3 - U2 < 0
hence for dU < 0, w < 0, heat flows out of the gas
iii. since heat flows out of the gas, entropy of the gas decreases
c. form 3 -> 1
i. volume is constant, so no work is done by the gas
ii. U(1) - U(3) = 3000 J > 0
hence
heat flows into the gas
iii. entropy of the gas increases
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