(a) The analytical formula for the variation of pressure with altitude.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 2, Problem 2.25P

To determine

(a)

The analytical formula for the variation of pressure with altitude.

Expert Solution

Answer to Problem 2.25P

Thus, the analytical formula for the variation of pressure with altitude is p=p0expgCRT01−eCz.

Explanation of Solution

Given information:

The approximate magnitude of acceleration due to gravity is 3.71 m/s2 . The atmosphere contains carbon dioxide. The average surface pressure is 700 Pa . The temperature distribution with respect to altitude from surface is T≈T0e−Cz . The magnitude of constant C is 1.3×10−5 m−1 . The magnitude of T0 is 250 K.

Write the expression for hydrostatic law.

dpdz=−ρg ...(I)

Here, the pressure of the fluid is p, the altitude of the fluid is z, the density of the fluid is ρ and the acceleration due to gravity for Mars is g.

Negative sign indicates fall in pressure with the rise in altitude.

Write the ideal gas equation.

p=ρRTρ=pRT

Here, the gas constant is R and the absolute temperature is T.

Substitute pRT for ρ in Equation (I).

dpdz=−pRTgdpp=−gRTdz ……. (II)

Write the expression for the variation of temperature.

T≈T0e−Cz ...(III)

Here, the constants are T0 and C.

Substitute T0e−Cz for T in Equation (II).

dpp=−gRT0e−Czdz

Integrate the above expression.

∫p0pdpp=−gRT0∫0zdze −Czlnpp0p=−gCRT0eCz−e0lnp−lnp0=gCRT01−eCzlnpp0=gCRT01−eCz

Here, the surface pressure is p0.

Further simplify the above expression.

p p 0=expgCR T 01− e Czp=p0expgCR T 01− e Cz

Conclusion:

The analytical formula for the variation of pressure with altitude is p=p0expgCRT01−eCz.

To determine

(b)

The altitude where the pressure on Mars has dropped to 1 Pa.

Expert Solution

Answer to Problem 2.25P

The altitude where the pressure on Mars has dropped to 1 Pa is 56.5 km.

Explanation of Solution

Given information:

The approximate magnitude of acceleration due to gravity is 3.71 m/s2 . The atmosphere contains carbon dioxide. The average surface pressure is 700 Pa . The temperature distribution with respect to altitude from surface is T≈T0e−Cz . The magnitude of constant C is 1.3×10−5 m−1 . The magnitude of T0 is 250 K.

Write the expression from the pressure distribution.

p=p0expgCRT01−eCz ...(IV)

Calculation:

Refer to Table A-4 to obtain gas constant as 189 m2/s2⋅K with respect to carbon dioxide.

Substitute 189 m2/s2⋅K for R, 1 Pa for p, 700 Pa for p0, 3.71 m/s2 for g, 1.3×10−5 m−1 for C and 250 K for T0 in Equation (IV).

1 Pa=700 Paexp 3.71 m/s2 1.3×10−5 m−1 189 m2/s2⋅K 250 K1− e 1.3×10−5 m−1 z ln1700= 3.71 m/s2 1.3×10−5 m−1 189 m2/s2⋅K 250 K1− e 1.3×10−5 m−1 z −6.551=6.041−e 1.3×10−5 m−1 z−1.0846=1−e1.3× 10 −5 m −1z

Further simplify the above expression.

−1.0846=1−e1.3× 10 −5 m −1ze1.3× 10 −5 m −1z=2.08461.3×10−5 m−1z=ln2.08461.3×10−5 m−1z=0.7345

Solve the above expression.

z=0.73451.3×10−5 m−1=56500 m1 km1000 m=56.5 km

Conclusion:

The altitude where the pressure on Mars has dropped to 1 Pa is 56.5 km.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

A start-up company wants to convert an ICE vehicle into an electric vehicle with the following specification. Power: 250 (HP) horsepower, (note: 1HP = 745 W) Range: 300-miles Fuel economy: 33.5 kilometers per gallon of gasoline. Efficiency of the ICE: 25% Energy Conversion: One gallon of gasoline at 100% efficiency is equal to 33.5 kWh/gallon). a)Calculate the EV consumption rate as Wh/km and find the total energy of the battery pack in KWh to replace the internal combustion engine. b)Design an 8-module battery pack for this full electric vehicle without compromising its range and performance (power). Use commercially available cylindrical cells lithium cell with 20Ah capacity and 3.125 V average voltage. Cell dimensions are 5cm diameter and 10 cm height. The electric motor requires 250 V input that will be provided directly from the battery pack, Report the configuration of each module in…

"11-17 The shaft shown in Figure P11-3 was designed in Problem 10-17. For the data in the row(s) assigned from Table P11-1, and the corresponding diameter of shaft found in Problem 10-17, design suitable bearings to support the load for at least 1E8 cycles at 1800 rpm. State all assumptions. (a) Using hydrodynamically lubricated bronze sleeve bearings with Ox = 15, 11d=0.75, and a clearance ratio of 0.001. ✓ ✓ cast-iron roller FIGURE P11-3 Shaft Design for Problems 11-17 b gear key assume bearings act as simple supports 11-19 The shaft shown in Figure P11-4 was designed in Problem 10-19. For the data in the row(s) assigned from Table P11-1, and the corresponding diameter of shaft found in Problem 10-19, design suitable bearings to support the load for at least 5E8 cycles at 1200 rpm. State all assumptions. (a) Using hydrodynamically lubricated bronze sleeve bearings with Oy = 40, 1/d=0.80, and a clearance ratio of 0.002 5. gear gear key FIGURE P11-4 Shaft Design for Problems 11-19 and…

For the frame below calculate the bending moment at point R. Take P=40 and note that this value is used for both the loads and the lengths of the members of the frame. 2.5P- A Q B R С 45 degrees ✗ ✗ P i 19 Кур -2P- 4PRN -P- -

Answer 2

Question

Chapter 2, Problem 2.25P

To determine

(a)

The analytical formula for the variation of pressure with altitude.

Expert Solution

Answer to Problem 2.25P

Thus, the analytical formula for the variation of pressure with altitude is p=p0expgCRT01−eCz.

Explanation of Solution

Given information:

The approximate magnitude of acceleration due to gravity is 3.71 m/s2 . The atmosphere contains carbon dioxide. The average surface pressure is 700 Pa . The temperature distribution with respect to altitude from surface is T≈T0e−Cz . The magnitude of constant C is 1.3×10−5 m−1 . The magnitude of T0 is 250 K.

Write the expression for hydrostatic law.

dpdz=−ρg ...(I)

Here, the pressure of the fluid is p, the altitude of the fluid is z, the density of the fluid is ρ and the acceleration due to gravity for Mars is g.

Negative sign indicates fall in pressure with the rise in altitude.

Write the ideal gas equation.

p=ρRTρ=pRT

Here, the gas constant is R and the absolute temperature is T.

Substitute pRT for ρ in Equation (I).

dpdz=−pRTgdpp=−gRTdz ……. (II)

Write the expression for the variation of temperature.

T≈T0e−Cz ...(III)

Here, the constants are T0 and C.

Substitute T0e−Cz for T in Equation (II).

dpp=−gRT0e−Czdz

Integrate the above expression.

∫p0pdpp=−gRT0∫0zdze −Czlnpp0p=−gCRT0eCz−e0lnp−lnp0=gCRT01−eCzlnpp0=gCRT01−eCz

Here, the surface pressure is p0.

Further simplify the above expression.

p p 0=expgCR T 01− e Czp=p0expgCR T 01− e Cz

Conclusion:

The analytical formula for the variation of pressure with altitude is p=p0expgCRT01−eCz.

To determine

(b)

The altitude where the pressure on Mars has dropped to 1 Pa.

Expert Solution

Answer to Problem 2.25P

The altitude where the pressure on Mars has dropped to 1 Pa is 56.5 km.

Explanation of Solution

Given information:

The approximate magnitude of acceleration due to gravity is 3.71 m/s2 . The atmosphere contains carbon dioxide. The average surface pressure is 700 Pa . The temperature distribution with respect to altitude from surface is T≈T0e−Cz . The magnitude of constant C is 1.3×10−5 m−1 . The magnitude of T0 is 250 K.

Write the expression from the pressure distribution.

p=p0expgCRT01−eCz ...(IV)

Calculation:

Refer to Table A-4 to obtain gas constant as 189 m2/s2⋅K with respect to carbon dioxide.

Substitute 189 m2/s2⋅K for R, 1 Pa for p, 700 Pa for p0, 3.71 m/s2 for g, 1.3×10−5 m−1 for C and 250 K for T0 in Equation (IV).

1 Pa=700 Paexp 3.71 m/s2 1.3×10−5 m−1 189 m2/s2⋅K 250 K1− e 1.3×10−5 m−1 z ln1700= 3.71 m/s2 1.3×10−5 m−1 189 m2/s2⋅K 250 K1− e 1.3×10−5 m−1 z −6.551=6.041−e 1.3×10−5 m−1 z−1.0846=1−e1.3× 10 −5 m −1z

Further simplify the above expression.

−1.0846=1−e1.3× 10 −5 m −1ze1.3× 10 −5 m −1z=2.08461.3×10−5 m−1z=ln2.08461.3×10−5 m−1z=0.7345

Solve the above expression.

z=0.73451.3×10−5 m−1=56500 m1 km1000 m=56.5 km

Conclusion:

The altitude where the pressure on Mars has dropped to 1 Pa is 56.5 km.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 2, Problem 2.25P

To determine

(a)

The analytical formula for the variation of pressure with altitude.

Expert Solution

Answer to Problem 2.25P

Thus, the analytical formula for the variation of pressure with altitude is p=p0expgCRT01−eCz.

Explanation of Solution

Given information:

The approximate magnitude of acceleration due to gravity is 3.71 m/s2 . The atmosphere contains carbon dioxide. The average surface pressure is 700 Pa . The temperature distribution with respect to altitude from surface is T≈T0e−Cz . The magnitude of constant C is 1.3×10−5 m−1 . The magnitude of T0 is 250 K.

Write the expression for hydrostatic law.

dpdz=−ρg ...(I)

Here, the pressure of the fluid is p, the altitude of the fluid is z, the density of the fluid is ρ and the acceleration due to gravity for Mars is g.

Negative sign indicates fall in pressure with the rise in altitude.

Write the ideal gas equation.

p=ρRTρ=pRT

Here, the gas constant is R and the absolute temperature is T.

Substitute pRT for ρ in Equation (I).

dpdz=−pRTgdpp=−gRTdz ……. (II)

Write the expression for the variation of temperature.

T≈T0e−Cz ...(III)

Here, the constants are T0 and C.

Substitute T0e−Cz for T in Equation (II).

dpp=−gRT0e−Czdz

Integrate the above expression.

∫p0pdpp=−gRT0∫0zdze −Czlnpp0p=−gCRT0eCz−e0lnp−lnp0=gCRT01−eCzlnpp0=gCRT01−eCz

Here, the surface pressure is p0.

Further simplify the above expression.

p p 0=expgCR T 01− e Czp=p0expgCR T 01− e Cz

Conclusion:

The analytical formula for the variation of pressure with altitude is p=p0expgCRT01−eCz.

To determine

(b)

The altitude where the pressure on Mars has dropped to 1 Pa.

Expert Solution

Answer to Problem 2.25P

The altitude where the pressure on Mars has dropped to 1 Pa is 56.5 km.

Explanation of Solution

Given information:

The approximate magnitude of acceleration due to gravity is 3.71 m/s2 . The atmosphere contains carbon dioxide. The average surface pressure is 700 Pa . The temperature distribution with respect to altitude from surface is T≈T0e−Cz . The magnitude of constant C is 1.3×10−5 m−1 . The magnitude of T0 is 250 K.

Write the expression from the pressure distribution.

p=p0expgCRT01−eCz ...(IV)

Calculation:

Refer to Table A-4 to obtain gas constant as 189 m2/s2⋅K with respect to carbon dioxide.

Substitute 189 m2/s2⋅K for R, 1 Pa for p, 700 Pa for p0, 3.71 m/s2 for g, 1.3×10−5 m−1 for C and 250 K for T0 in Equation (IV).

1 Pa=700 Paexp 3.71 m/s2 1.3×10−5 m−1 189 m2/s2⋅K 250 K1− e 1.3×10−5 m−1 z ln1700= 3.71 m/s2 1.3×10−5 m−1 189 m2/s2⋅K 250 K1− e 1.3×10−5 m−1 z −6.551=6.041−e 1.3×10−5 m−1 z−1.0846=1−e1.3× 10 −5 m −1z

Further simplify the above expression.

−1.0846=1−e1.3× 10 −5 m −1ze1.3× 10 −5 m −1z=2.08461.3×10−5 m−1z=ln2.08461.3×10−5 m−1z=0.7345

Solve the above expression.

z=0.73451.3×10−5 m−1=56500 m1 km1000 m=56.5 km

Conclusion:

The altitude where the pressure on Mars has dropped to 1 Pa is 56.5 km.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 2, Problem 2.25P

To determine

(a)

The analytical formula for the variation of pressure with altitude.

Expert Solution

Answer to Problem 2.25P

Thus, the analytical formula for the variation of pressure with altitude is p=p0expgCRT01−eCz.

Explanation of Solution

Given information:

The approximate magnitude of acceleration due to gravity is 3.71 m/s2 . The atmosphere contains carbon dioxide. The average surface pressure is 700 Pa . The temperature distribution with respect to altitude from surface is T≈T0e−Cz . The magnitude of constant C is 1.3×10−5 m−1 . The magnitude of T0 is 250 K.

Write the expression for hydrostatic law.

dpdz=−ρg ...(I)

Here, the pressure of the fluid is p, the altitude of the fluid is z, the density of the fluid is ρ and the acceleration due to gravity for Mars is g.

Negative sign indicates fall in pressure with the rise in altitude.

Write the ideal gas equation.

p=ρRTρ=pRT

Here, the gas constant is R and the absolute temperature is T.

Substitute pRT for ρ in Equation (I).

dpdz=−pRTgdpp=−gRTdz ……. (II)

Write the expression for the variation of temperature.

T≈T0e−Cz ...(III)

Here, the constants are T0 and C.

Substitute T0e−Cz for T in Equation (II).

dpp=−gRT0e−Czdz

Integrate the above expression.

∫p0pdpp=−gRT0∫0zdze −Czlnpp0p=−gCRT0eCz−e0lnp−lnp0=gCRT01−eCzlnpp0=gCRT01−eCz

Here, the surface pressure is p0.

Further simplify the above expression.

p p 0=expgCR T 01− e Czp=p0expgCR T 01− e Cz

Conclusion:

The analytical formula for the variation of pressure with altitude is p=p0expgCRT01−eCz.

To determine

(b)

The altitude where the pressure on Mars has dropped to 1 Pa.

Expert Solution

Answer to Problem 2.25P

The altitude where the pressure on Mars has dropped to 1 Pa is 56.5 km.

Explanation of Solution

Given information:

The approximate magnitude of acceleration due to gravity is 3.71 m/s2 . The atmosphere contains carbon dioxide. The average surface pressure is 700 Pa . The temperature distribution with respect to altitude from surface is T≈T0e−Cz . The magnitude of constant C is 1.3×10−5 m−1 . The magnitude of T0 is 250 K.

Write the expression from the pressure distribution.

p=p0expgCRT01−eCz ...(IV)

Calculation:

Refer to Table A-4 to obtain gas constant as 189 m2/s2⋅K with respect to carbon dioxide.

Substitute 189 m2/s2⋅K for R, 1 Pa for p, 700 Pa for p0, 3.71 m/s2 for g, 1.3×10−5 m−1 for C and 250 K for T0 in Equation (IV).

1 Pa=700 Paexp 3.71 m/s2 1.3×10−5 m−1 189 m2/s2⋅K 250 K1− e 1.3×10−5 m−1 z ln1700= 3.71 m/s2 1.3×10−5 m−1 189 m2/s2⋅K 250 K1− e 1.3×10−5 m−1 z −6.551=6.041−e 1.3×10−5 m−1 z−1.0846=1−e1.3× 10 −5 m −1z

Further simplify the above expression.

−1.0846=1−e1.3× 10 −5 m −1ze1.3× 10 −5 m −1z=2.08461.3×10−5 m−1z=ln2.08461.3×10−5 m−1z=0.7345

Solve the above expression.

z=0.73451.3×10−5 m−1=56500 m1 km1000 m=56.5 km

Conclusion:

The altitude where the pressure on Mars has dropped to 1 Pa is 56.5 km.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 2, Problem 2.25P

To determine

(a)

The analytical formula for the variation of pressure with altitude.

Expert Solution

Answer to Problem 2.25P

Thus, the analytical formula for the variation of pressure with altitude is p=p0expgCRT01−eCz.

Explanation of Solution

Given information:

The approximate magnitude of acceleration due to gravity is 3.71 m/s2 . The atmosphere contains carbon dioxide. The average surface pressure is 700 Pa . The temperature distribution with respect to altitude from surface is T≈T0e−Cz . The magnitude of constant C is 1.3×10−5 m−1 . The magnitude of T0 is 250 K.

Write the expression for hydrostatic law.

dpdz=−ρg ...(I)

Here, the pressure of the fluid is p, the altitude of the fluid is z, the density of the fluid is ρ and the acceleration due to gravity for Mars is g.

Negative sign indicates fall in pressure with the rise in altitude.

Write the ideal gas equation.

p=ρRTρ=pRT

Here, the gas constant is R and the absolute temperature is T.

Substitute pRT for ρ in Equation (I).

dpdz=−pRTgdpp=−gRTdz ……. (II)

Write the expression for the variation of temperature.

T≈T0e−Cz ...(III)

Here, the constants are T0 and C.

Substitute T0e−Cz for T in Equation (II).

dpp=−gRT0e−Czdz

Integrate the above expression.

∫p0pdpp=−gRT0∫0zdze −Czlnpp0p=−gCRT0eCz−e0lnp−lnp0=gCRT01−eCzlnpp0=gCRT01−eCz

Here, the surface pressure is p0.

Further simplify the above expression.

p p 0=expgCR T 01− e Czp=p0expgCR T 01− e Cz

Conclusion:

The analytical formula for the variation of pressure with altitude is p=p0expgCRT01−eCz.

To determine

(b)

The altitude where the pressure on Mars has dropped to 1 Pa.

Expert Solution

Answer to Problem 2.25P

The altitude where the pressure on Mars has dropped to 1 Pa is 56.5 km.

Explanation of Solution

Given information:

The approximate magnitude of acceleration due to gravity is 3.71 m/s2 . The atmosphere contains carbon dioxide. The average surface pressure is 700 Pa . The temperature distribution with respect to altitude from surface is T≈T0e−Cz . The magnitude of constant C is 1.3×10−5 m−1 . The magnitude of T0 is 250 K.

Write the expression from the pressure distribution.

p=p0expgCRT01−eCz ...(IV)

Calculation:

Refer to Table A-4 to obtain gas constant as 189 m2/s2⋅K with respect to carbon dioxide.

Substitute 189 m2/s2⋅K for R, 1 Pa for p, 700 Pa for p0, 3.71 m/s2 for g, 1.3×10−5 m−1 for C and 250 K for T0 in Equation (IV).

1 Pa=700 Paexp 3.71 m/s2 1.3×10−5 m−1 189 m2/s2⋅K 250 K1− e 1.3×10−5 m−1 z ln1700= 3.71 m/s2 1.3×10−5 m−1 189 m2/s2⋅K 250 K1− e 1.3×10−5 m−1 z −6.551=6.041−e 1.3×10−5 m−1 z−1.0846=1−e1.3× 10 −5 m −1z

Further simplify the above expression.

−1.0846=1−e1.3× 10 −5 m −1ze1.3× 10 −5 m −1z=2.08461.3×10−5 m−1z=ln2.08461.3×10−5 m−1z=0.7345

Solve the above expression.

z=0.73451.3×10−5 m−1=56500 m1 km1000 m=56.5 km

Conclusion:

The altitude where the pressure on Mars has dropped to 1 Pa is 56.5 km.

Answer 6

Chapter 2, Problem 2.25P

To determine

(a)

The analytical formula for the variation of pressure with altitude.

Expert Solution

Answer to Problem 2.25P

Thus, the analytical formula for the variation of pressure with altitude is p=p0expgCRT01−eCz.

Explanation of Solution

Given information:

The approximate magnitude of acceleration due to gravity is 3.71 m/s2 . The atmosphere contains carbon dioxide. The average surface pressure is 700 Pa . The temperature distribution with respect to altitude from surface is T≈T0e−Cz . The magnitude of constant C is 1.3×10−5 m−1 . The magnitude of T0 is 250 K.

Write the expression for hydrostatic law.

dpdz=−ρg ...(I)

Here, the pressure of the fluid is p, the altitude of the fluid is z, the density of the fluid is ρ and the acceleration due to gravity for Mars is g.

Negative sign indicates fall in pressure with the rise in altitude.

Write the ideal gas equation.

p=ρRTρ=pRT

Here, the gas constant is R and the absolute temperature is T.

Substitute pRT for ρ in Equation (I).

dpdz=−pRTgdpp=−gRTdz ……. (II)

Write the expression for the variation of temperature.

T≈T0e−Cz ...(III)

Here, the constants are T0 and C.

Substitute T0e−Cz for T in Equation (II).

dpp=−gRT0e−Czdz

Integrate the above expression.

∫p0pdpp=−gRT0∫0zdze −Czlnpp0p=−gCRT0eCz−e0lnp−lnp0=gCRT01−eCzlnpp0=gCRT01−eCz

Here, the surface pressure is p0.

Further simplify the above expression.

p p 0=expgCR T 01− e Czp=p0expgCR T 01− e Cz

Conclusion:

The analytical formula for the variation of pressure with altitude is p=p0expgCRT01−eCz.

Answer 7

Chapter 2, Problem 2.25P

To determine

(a)

The analytical formula for the variation of pressure with altitude.

Answer 8

Expert Solution

Answer to Problem 2.25P

Thus, the analytical formula for the variation of pressure with altitude is p=p0expgCRT01−eCz.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

The approximate magnitude of acceleration due to gravity is 3.71 m/s2 . The atmosphere contains carbon dioxide. The average surface pressure is 700 Pa . The temperature distribution with respect to altitude from surface is T≈T0e−Cz . The magnitude of constant C is 1.3×10−5 m−1 . The magnitude of T0 is 250 K.

Write the expression for hydrostatic law.

dpdz=−ρg ...(I)

Here, the pressure of the fluid is p, the altitude of the fluid is z, the density of the fluid is ρ and the acceleration due to gravity for Mars is g.

Negative sign indicates fall in pressure with the rise in altitude.

Write the ideal gas equation.

p=ρRTρ=pRT

Here, the gas constant is R and the absolute temperature is T.

Substitute pRT for ρ in Equation (I).

dpdz=−pRTgdpp=−gRTdz ……. (II)

Write the expression for the variation of temperature.

T≈T0e−Cz ...(III)

Here, the constants are T0 and C.

Substitute T0e−Cz for T in Equation (II).

dpp=−gRT0e−Czdz

Integrate the above expression.

∫p0pdpp=−gRT0∫0zdze −Czlnpp0p=−gCRT0eCz−e0lnp−lnp0=gCRT01−eCzlnpp0=gCRT01−eCz

Here, the surface pressure is p0.

Further simplify the above expression.

p p 0=expgCR T 01− e Czp=p0expgCR T 01− e Cz

Conclusion:

The analytical formula for the variation of pressure with altitude is p=p0expgCRT01−eCz.

Answer 13

To determine

(b)

The altitude where the pressure on Mars has dropped to 1 Pa.

Expert Solution

Answer to Problem 2.25P

The altitude where the pressure on Mars has dropped to 1 Pa is 56.5 km.

Explanation of Solution

Given information:

The approximate magnitude of acceleration due to gravity is 3.71 m/s2 . The atmosphere contains carbon dioxide. The average surface pressure is 700 Pa . The temperature distribution with respect to altitude from surface is T≈T0e−Cz . The magnitude of constant C is 1.3×10−5 m−1 . The magnitude of T0 is 250 K.

Write the expression from the pressure distribution.

p=p0expgCRT01−eCz ...(IV)

Calculation:

Refer to Table A-4 to obtain gas constant as 189 m2/s2⋅K with respect to carbon dioxide.

Substitute 189 m2/s2⋅K for R, 1 Pa for p, 700 Pa for p0, 3.71 m/s2 for g, 1.3×10−5 m−1 for C and 250 K for T0 in Equation (IV).

1 Pa=700 Paexp 3.71 m/s2 1.3×10−5 m−1 189 m2/s2⋅K 250 K1− e 1.3×10−5 m−1 z ln1700= 3.71 m/s2 1.3×10−5 m−1 189 m2/s2⋅K 250 K1− e 1.3×10−5 m−1 z −6.551=6.041−e 1.3×10−5 m−1 z−1.0846=1−e1.3× 10 −5 m −1z

Further simplify the above expression.

−1.0846=1−e1.3× 10 −5 m −1ze1.3× 10 −5 m −1z=2.08461.3×10−5 m−1z=ln2.08461.3×10−5 m−1z=0.7345

Solve the above expression.

z=0.73451.3×10−5 m−1=56500 m1 km1000 m=56.5 km

Conclusion:

The altitude where the pressure on Mars has dropped to 1 Pa is 56.5 km.

Answer 14

To determine

(b)

The altitude where the pressure on Mars has dropped to 1 Pa.

Answer 15

Expert Solution

Answer to Problem 2.25P

The altitude where the pressure on Mars has dropped to 1 Pa is 56.5 km.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:

The approximate magnitude of acceleration due to gravity is 3.71 m/s2 . The atmosphere contains carbon dioxide. The average surface pressure is 700 Pa . The temperature distribution with respect to altitude from surface is T≈T0e−Cz . The magnitude of constant C is 1.3×10−5 m−1 . The magnitude of T0 is 250 K.

Write the expression from the pressure distribution.

p=p0expgCRT01−eCz ...(IV)

Calculation:

Refer to Table A-4 to obtain gas constant as 189 m2/s2⋅K with respect to carbon dioxide.

Substitute 189 m2/s2⋅K for R, 1 Pa for p, 700 Pa for p0, 3.71 m/s2 for g, 1.3×10−5 m−1 for C and 250 K for T0 in Equation (IV).

1 Pa=700 Paexp 3.71 m/s2 1.3×10−5 m−1 189 m2/s2⋅K 250 K1− e 1.3×10−5 m−1 z ln1700= 3.71 m/s2 1.3×10−5 m−1 189 m2/s2⋅K 250 K1− e 1.3×10−5 m−1 z −6.551=6.041−e 1.3×10−5 m−1 z−1.0846=1−e1.3× 10 −5 m −1z