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

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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.

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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.

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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.

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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.

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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