A 4.00 L sample of a diatomic ideal gas with specific heat ratio 1.40, confined

Question

A 4.00 L sample of a diatomic ideal gas with specific heat ratio 1.40, confined to a cylinder, is carried through a closed cycle. The gas is initially at 1.00 atm and at 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressures. Finally, the gas is compressed isobarically to its original volume. Draw a PV diagram of this cycle. (Do this on paper. Your instructor may ask you to turn in this work.) Determine the volume of the gas at the end of the adiabatic expansion. L Find the temperature of the gas at the start of the adiabatic expansion. K Find the temperature at the end of the cycle. K What was the net work done on the gas for this cycle? J

Explanation / Answer

The cycle is characterized by 3 points in the pV plane; each point corresponds to a definite "state" of gas. In state 1,
p1 = 1 atm
V1 = 4 l
T1 = 300 K.

Point 2 is just 2 units above point 1 in the pV plane. From given data,
p2 = 3 atm
V2 = V1 = 4 l
T2 = ?

Point 3 is located horizontally to the left of point 1; it is known only that p3 = p1:
p3 = p1 = 1 atm
V3 = ?
T3 = ?

a) V3 is to be determined. Previous volume is V2. For an adiabatic expansion,

p3 V3^ = p2 V2^
V3/V2 = (p2/p3)^1/
V3 = V2 (p2/p3)^1/.

But V2 = V1 = 4 l, and p3 = p1 = 1 atm, so

V3 = 4 (3/1)^1/1.4 = 8.7672 l.

b) T2 is required. Previous temperature is T1 = 300 K. For this isometric process,

p2/T2 = p1/T1
T2 = T1 · p2/p1 = 300 × 3 = 900 K.

c) I'm not clear as to what is meant by "temperature at the end of the cycle". For me, the cycle ends when the gas returns to its initial state (state 1), so the temperature is clearly 300 K. However, should T3 be meant instead, this can be calculated from

V3/T3 = V1/T1
T3 = T1 · V3/V1 = 300 × 8.7672/4 = 657.55 K.

d) Again, this question can be construed in two ways. Certain amount of work is done on gas in order to compress it again to original volume (4 l). This is easily computed as W = p V = 1 (8.7672 4) = 4.7672 atm·l = 483 J. This is the work on gas per cycle.

On the other hand, as the gas expands adiabatically, it does mechanical work equal to 8.082 atm·l*, or 819 J. This exceeds the amount of work done on gas; thus, another possible answer is 483 819 = 336 J, which means 336 net J mechanical work done by gas per cycle.

* W = (p2 V2 p3 V3) / ( 1), work done by gas on an adiabatic process.

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