# Physical Exercises

Classified in Physics

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Six 19.4 moles of ideal gas are in a cylinder fitted at one end with a movable piston. The initial gas temperature is 27.0 ° C and the pressure is constant. As part of a proposed machine design, calculate the final temperature of the gas once it has made 1.75x10 ³ J of work.

n = 6 moles isobaric process

T? = 27 ° C

P = CTE.

W = 1.75 * 10 ³ J

PV = nRT

V? = Vf =

W = P (-)

Pf = P: = P

W = (nRTf - nRT?)

W + nRT? = NRTf

Tf =

Tf =

19.8

1 4 2 3 1? Explain.

1.3 Isobaric Process

3.2 isochoric

2.4 Isobaric

4.1 isochoric

a) Wt = W1-3 + W3-2 + W2-4 + W4-1

Wt = P1 (V3-V1) + P2 (V4-V2)

But

V1 = V4 and V2 = V3 replacing

Wt = P1 (V2-V1) + P2 (V1-V2) = P1 (V2-V1)-P2 (V2-V1)

Wt = (P1-P2) (V2-V1)

b * h

b) 1.4 isochoric process

3.1 Isobaric

2.3 isochoric

4.2 Isobaric

Wt = W1-4 + W4-2 + W2-3 + W3-1

Wt = P1 (V3-V1) + P2 (V4-V2

V1 = V4 and V2 = V3

Wt = P1 (V3-Vl) + P2 (V4-V2) =-P1 (V3 + V1) + P2 (V4-V2)

Wt = (-p1 + p2) (V2-V1)

Explanation

V

a) The heat depends on the path absorbs heat Q = + is known through the path a - b gives more heat than b - a

b) When is a closed cycle ÄU = 0 Q = 7200J

ÄU QW = Q = W

W = 7200J The system does work on its surroundings

c) ab positive (+) and ba negative (-)

Da negative fall out nothing but heat

Q = apparent heat-7200J

Q = -7200 = 7200 J of heat Magnitude calort absorbs heat from the didteme

19.21 In an experiment to simulate conditions inside a car engine, 645 J of heat is transferred to 0185 moles of air contained in a cylinder whose volume is 40.0 cm ³. At first air at a pressure of 3.00 × 10? Pa and a temperature of 780 K. a) If the cylinder volume is fixed, what the air temperature reaches final? Assume the air is almost pure nitrogen and use the data in Table 19.1 but the pressure is not low. Draw a graph pV for this process. b) Calculate the final temperature of the air if allowed to increase the cylinder volume while the pressure remains constant. Draw a graph pV for this process.

Q = 645J

n = 0185 mol

V = 40.0 cm ³

P? = 3.00x10? Pa

T? = 780k

a) Tf =? When the volume is constant

Cv = 20.76 J / mol.K

Q = nCvÄT

(Tf-T?) =

Tf =

Tf =

Tf =

Tf = 167.97K + 780k

b) Cp = 29.7J/molK

Q = nCpÄT

AT =

(Tf-T?) =

Tf =

Tf =

19.27 moles 0.150-temperature ideal gas is held constant at 77.0 º C while its volume is reduced to 25.0% of its initial volume. The initial gas pressure is 1.25 atm. a) Determine the work done by the gas. B) Determine the internal energy change. c) Is the gas exchanges heat with its environment? If so, how much? The gas absorbs or releases heat?

T = 77.0 ° C CTE. T = 77 ° C +273016 = 350.16K

n = 0.150mol

Vf = 25%-V?

P? = 1.25atm

ÄU = 0

R = 8.314J/molK

a)

Q = W = nRTLn (Vf / V?)

Pv = nRT

V? =

V? =

V? 4.31lt = 4.31 * 10 ¯ ³ m³

Vf = 4.31-75%

1.08lt Vf = 1.08 * 10 ¯ ³ m³

W = nRTLn (Vf / V?)

W = 0.150mol (8.314J/molK) (350.16K) Ln (8.62 * 10 ¯? M³ / 3.45 * 10 ¯ ³ m³)

b)

AU = 0 because it is an isothermal process

c)

Yes, the gas exchanges heat with its surroundings

Q = W = AU

W Q = W <0

19.30 A cylinder contains 0.250 moles of carbon dioxide (CO?) Gas at a temperature of 27.0 ° C. The cylinder has a piston without friction, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its temperature rises to 127.0 ° C assume that the CO? can be treated as ideal gas. a) Draw a graph pV for this process. b) How much work making the gas in this process? c) On what the job is done? d) When changing the internal energy of gas? e) How much heat is supplied to gas? f) How much work would have been incurred if the pressure would have been 0.50 atm?

n = 0.250mol

T? = 27.0 ° C 300.16K

P = 1.00 atm CTE.

Tf = 127.0 ° C 400.16K

a) Grafica Pv

b)

W = P (Vf-V?)

Pv = nRT

V? =

V? =

V? 6.16Lt = 6.16 * 10 ¯ ³ m³

Vf =

Vf =

8.21Lt Vf = 8.21 * 10 ¯ ³ m³

W = P (Vf-V?)

W = 1,013 * 10? Pa (8.21 * 10 ³ m³ ¯ ¯ ³ -6.16 * 10 m³)

c) The

d) AU = nCvÄT

At T = Tf-T? = 400.16K-300.16K = 100K

CO? Cv = 28.46J/molK

Au = (0.250mol) (28.46J/molK) (100K)

e) Q =?

QW = AU

Q = AU + W

Q = 711.5J +207.7

f)

19.34 Two moles of carbon monoxide (CO) are at a pressure of 1.2 atm and occupy a volume of 30 liters. Then the gas is compressed adiabatically? of that volume. Suppose the gas is ideal behavior. What is the change in internal energy? "The internal energy increases or decreases? Does the temperature of the gas increases or decreases during the process? Explain.

n = 2mol

P? = 1.2 atm

V? = 30 Lt

Vf = V? = Vf = 10 Lt

AU =?

Q = 0

Ã =

=

Increases because work ÄU gas> 0

The temperature increases because UA is increasing because it is an ideal gas.

19.39 An amount of sulfur dioxide (SO?) Gas occupies a volume of 5.00 × 10 ¯ ³ m³ at a pressure of 1.10 × 10? Pa. The gas expands adiabatically to a volume of 1.00 × 10 ¯ ² m³. Suppose the gas is ideal behavior. a) Calculate the final pressure of gas. (Hint: See Table 19.1. B) How much work carried out by the gas on its surroundings? c) Determine the ratio final temperature: initial temperature of the gas.

V? = 5.00 * 10 m³

P: = 1.10 * 10Pa

ã = 1.29

V? = 1.00 * 10 ¯ ² m³

a) P: =?

P: =

P: = Pa 44.98.59

P: = 4.5 * 10? Pa

b) W =?

W =

W =

c) T:: T?

19.44 A thermodynamic system is carried from state to state c in Figure 19.29 along the path abc, the work W done by the system is 450 J. For the path abc, W is 120 J. internal energies of the four states shown in the figure are: Ua = 150 J = 240 J Ub, Uc and Ud = 680 J = 330 J. Calculate the heat flow Q for each of the four processes: ab, bc, ad and dc. In each process, does the system absorbs or releases heat?

Wabc = 450J

WADC = 120J

Ua = 150J

Ub = 240J

Uc = 680J

Ud = 330J

QW = AU

Q = AU + W

Q = (Uf-U?) + W

Heat flow in ab

Q = (240J-150J)

Q (240J1505J + W

Flow DECALE bc

Q = (680J-240J) +450 J

Heat flow in a1

Q = (330J-150J) 120J

Q = 300J

Heat flow in dc

Q = (680J-330J) + W

19.47 Two moles of a monatomic gas with ideal behavior cycled abc. In a complete cycle, 800 J of heat out of gas. The process is carried out ab at constant pressure, and the bc, at constant volume. B statements have temperatures Ta = 200 K and Tb = 300 K. a) Draw a graph for the cycle pV. b) How much work W is done in the process ca?

Q =- 800J n = 2mol

Ta = 200K R = 8.314J/mol.K

Tb = 300K

a)

ab = bc = Constant Pressure Constant Volume

b) QW = AU

QW = (Uf-U?) U? = Uf

QW = 0

W Q = -800 J

Sale system heat and work is done

Q <0 and W <0

Wab = P (Vf-V?)

Wab = PV

Wab = nR T

Wab = (2mol) (8.314J/molK) (300K-200K)

Wab = 1662.8J

Wbc Wtotal = Wab + + WCA

Wca = Wtotal-Wab

Wca-800J =- 1662.8J

n = 6 moles isobaric process

T? = 27 ° C

P = CTE.

W = 1.75 * 10 ³ J

PV = nRT

V? = Vf =

W = P (-)

Pf = P: = P

W = (nRTf - nRT?)

W + nRT? = NRTf

Tf =

Tf =

**Tf = 335.24K**19.8

**Work done in a cyclic****process.**a) In Figure 19.8a, consider the closed loop 1 3 2 4 1. This is a cyclical process in which the initial and final states are the same. Calculate the total work done by the system in this process and show that equals the area enclosed by the cycle. b) What is the relationship between the work done by the process of part (a) and if made through the loop in the opposite direction,1 4 2 3 1? Explain.

1.3 Isobaric Process

3.2 isochoric

2.4 Isobaric

4.1 isochoric

a) Wt = W1-3 + W3-2 + W2-4 + W4-1

Wt = P1 (V3-V1) + P2 (V4-V2)

But

V1 = V4 and V2 = V3 replacing

Wt = P1 (V2-V1) + P2 (V1-V2) = P1 (V2-V1)-P2 (V2-V1)

Wt = (P1-P2) (V2-V1)

b * h

b) 1.4 isochoric process

3.1 Isobaric

2.3 isochoric

4.2 Isobaric

Wt = W1-4 + W4-2 + W2-3 + W3-1

Wt = P1 (V3-V1) + P2 (V4-V2

V1 = V4 and V2 = V3

Wt = P1 (V3-Vl) + P2 (V4-V2) =-P1 (V3 + V1) + P2 (V4-V2)

Wt = (-p1 + p2) (V2-V1)

Explanation

**The area is the same as what changes is the sign**

19.17 A system is carried by the cycle of Figure 19.23, the state a to b and back to a. The vabsolute alue of heat transfer during a cycle is of 7200 J. a) Does the system absorbs or releases heat when it through the loop in the direction indicated in the figure? How do you know? b) What makes the system work W into a cycle? c) If the system through the loop in a counterclockwise direction, "absorbs or releases heat in the cycle? How big is the heat absorbed or evolved in a counterclockwise cycle?V

a) The heat depends on the path absorbs heat Q = + is known through the path a - b gives more heat than b - a

b) When is a closed cycle ÄU = 0 Q = 7200J

ÄU QW = Q = W

W = 7200J The system does work on its surroundings

c) ab positive (+) and ba negative (-)

Da negative fall out nothing but heat

Q = apparent heat-7200J

Q = -7200 = 7200 J of heat Magnitude calort absorbs heat from the didteme

19.21 In an experiment to simulate conditions inside a car engine, 645 J of heat is transferred to 0185 moles of air contained in a cylinder whose volume is 40.0 cm ³. At first air at a pressure of 3.00 × 10? Pa and a temperature of 780 K. a) If the cylinder volume is fixed, what the air temperature reaches final? Assume the air is almost pure nitrogen and use the data in Table 19.1 but the pressure is not low. Draw a graph pV for this process. b) Calculate the final temperature of the air if allowed to increase the cylinder volume while the pressure remains constant. Draw a graph pV for this process.

Q = 645J

n = 0185 mol

V = 40.0 cm ³

P? = 3.00x10? Pa

T? = 780k

a) Tf =? When the volume is constant

Cv = 20.76 J / mol.K

Q = nCvÄT

(Tf-T?) =

Tf =

Tf =

Tf =

Tf = 167.97K + 780k

**Tf = 947.97K**

b) Cp = 29.7J/molK

Q = nCpÄT

AT =

(Tf-T?) =

Tf =

Tf =

**Tf = 899.89 K**19.27 moles 0.150-temperature ideal gas is held constant at 77.0 º C while its volume is reduced to 25.0% of its initial volume. The initial gas pressure is 1.25 atm. a) Determine the work done by the gas. B) Determine the internal energy change. c) Is the gas exchanges heat with its environment? If so, how much? The gas absorbs or releases heat?

T = 77.0 ° C CTE. T = 77 ° C +273016 = 350.16K

n = 0.150mol

Vf = 25%-V?

P? = 1.25atm

ÄU = 0

R = 8.314J/molK

a)

Q = W = nRTLn (Vf / V?)

Pv = nRT

V? =

V? =

V? 4.31lt = 4.31 * 10 ¯ ³ m³

Vf = 4.31-75%

1.08lt Vf = 1.08 * 10 ¯ ³ m³

W = nRTLn (Vf / V?)

W = 0.150mol (8.314J/molK) (350.16K) Ln (8.62 * 10 ¯? M³ / 3.45 * 10 ¯ ³ m³)

**W =- 605.63****The surroundings do work on the system.**b)

AU = 0 because it is an isothermal process

c)

Yes, the gas exchanges heat with its surroundings

Q = W = AU

W Q = W <0

**Q =- 605.63****The heat released is 605.63J**19.30 A cylinder contains 0.250 moles of carbon dioxide (CO?) Gas at a temperature of 27.0 ° C. The cylinder has a piston without friction, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its temperature rises to 127.0 ° C assume that the CO? can be treated as ideal gas. a) Draw a graph pV for this process. b) How much work making the gas in this process? c) On what the job is done? d) When changing the internal energy of gas? e) How much heat is supplied to gas? f) How much work would have been incurred if the pressure would have been 0.50 atm?

n = 0.250mol

T? = 27.0 ° C 300.16K

P = 1.00 atm CTE.

Tf = 127.0 ° C 400.16K

a) Grafica Pv

b)

W = P (Vf-V?)

Pv = nRT

V? =

V? =

V? 6.16Lt = 6.16 * 10 ¯ ³ m³

Vf =

Vf =

8.21Lt Vf = 8.21 * 10 ¯ ³ m³

W = P (Vf-V?)

W = 1,013 * 10? Pa (8.21 * 10 ³ m³ ¯ ¯ ³ -6.16 * 10 m³)

**W = 207.7 J**c) The

**work is in the piston**d) AU = nCvÄT

At T = Tf-T? = 400.16K-300.16K = 100K

CO? Cv = 28.46J/molK

Au = (0.250mol) (28.46J/molK) (100K)

**AU = 711.5J**e) Q =?

QW = AU

Q = AU + W

Q = 711.5J +207.7

**Q = 919.2J**f)

**The work would be the same bone 207.85J**19.34 Two moles of carbon monoxide (CO) are at a pressure of 1.2 atm and occupy a volume of 30 liters. Then the gas is compressed adiabatically? of that volume. Suppose the gas is ideal behavior. What is the change in internal energy? "The internal energy increases or decreases? Does the temperature of the gas increases or decreases during the process? Explain.

n = 2mol

P? = 1.2 atm

V? = 30 Lt

Vf = V? = Vf = 10 Lt

AU =?

Q = 0

Ã =

=

Increases because work ÄU gas> 0

The temperature increases because UA is increasing because it is an ideal gas.

19.39 An amount of sulfur dioxide (SO?) Gas occupies a volume of 5.00 × 10 ¯ ³ m³ at a pressure of 1.10 × 10? Pa. The gas expands adiabatically to a volume of 1.00 × 10 ¯ ² m³. Suppose the gas is ideal behavior. a) Calculate the final pressure of gas. (Hint: See Table 19.1. B) How much work carried out by the gas on its surroundings? c) Determine the ratio final temperature: initial temperature of the gas.

V? = 5.00 * 10 m³

P: = 1.10 * 10Pa

ã = 1.29

V? = 1.00 * 10 ¯ ² m³

a) P: =?

P: =

P: = Pa 44.98.59

P: = 4.5 * 10? Pa

b) W =?

W =

W =

**W = 334.83 J**c) T:: T?

**The final temperature (T?) Is going down**19.44 A thermodynamic system is carried from state to state c in Figure 19.29 along the path abc, the work W done by the system is 450 J. For the path abc, W is 120 J. internal energies of the four states shown in the figure are: Ua = 150 J = 240 J Ub, Uc and Ud = 680 J = 330 J. Calculate the heat flow Q for each of the four processes: ab, bc, ad and dc. In each process, does the system absorbs or releases heat?

Wabc = 450J

WADC = 120J

Ua = 150J

Ub = 240J

Uc = 680J

Ud = 330J

QW = AU

Q = AU + W

Q = (Uf-U?) + W

Heat flow in ab

Q = (240J-150J)

Q (240J1505J + W

**Q = 90J**Flow DECALE bc

Q = (680J-240J) +450 J

**Q = 890J**Heat flow in a1

Q = (330J-150J) 120J

Q = 300J

Heat flow in dc

Q = (680J-330J) + W

**Q = 350J**

The system receives heat in each process.The system receives heat in each process.

19.47 Two moles of a monatomic gas with ideal behavior cycled abc. In a complete cycle, 800 J of heat out of gas. The process is carried out ab at constant pressure, and the bc, at constant volume. B statements have temperatures Ta = 200 K and Tb = 300 K. a) Draw a graph for the cycle pV. b) How much work W is done in the process ca?

Q =- 800J n = 2mol

Ta = 200K R = 8.314J/mol.K

Tb = 300K

a)

ab = bc = Constant Pressure Constant Volume

b) QW = AU

QW = (Uf-U?) U? = Uf

QW = 0

W Q = -800 J

Sale system heat and work is done

Q <0 and W <0

Wab = P (Vf-V?)

Wab = PV

Wab = nR T

Wab = (2mol) (8.314J/molK) (300K-200K)

Wab = 1662.8J

Wbc Wtotal = Wab + + WCA

Wca = Wtotal-Wab

Wca-800J =- 1662.8J

**Wca =- 2462.8J**