The rms speed of the H 2 molecules.

Question

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Answer 1

Question

Chapter 17, Problem 26P

(a)

To determine

The rms speed of the $H2$ molecules.

(a)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$ ........ (1)

Here, $vrms$ is the root mean-square speed of the molecule, $R$ is the gas constant, $T$ is the temperature and $M$ is the molar mass.

Write the equation for the conversion of Celsius to kelvin.

$T(°K)=T(°C)+273.15$ ........ (2)

$T(°C)$ Is the temperature in Celsius and $T(°K)$ is the temperature in kelvin.

Calculation:

Substitute $0 °C$ for $T(°C)$ in equation (2).

$T(°K)=(0 °C)+273.15T(°K)=273.15 K$

Substitute $273.15 K$ for $T$ , $2×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K )( 273.15 K ) 2× 10 −3 kg/ mol vrms=1.84 Km/s$

Conclusion:

The rms speed of the $H2$ molecules is $1.84 Km/s$ .

(b)

To determine

The rms speed of $O2$ molecules.

(b)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$

Calculation:

Substitute $273.15 K$ for $T$ , $32×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K −1 )( 273.15 K ) 32× 10 −3 kg/ mol vrms= 461 m/s$

Conclusion:

The root mean square speed for the oxygen molecule is $461 m/s$ .

(c)

To determine

The root mean square speed of the $CO2$ molecule.

(c)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$

Calculation:

Substitute $273.15 K$ for $T$ , $44×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K −1 )( 273.15 K ) 44× 10 −3 kg/ mol vrms= 393 m/s$

Conclusion:

The root mean square speed for the $CO2$ molecule is $393 m/s$ .

(d)

To determine

Whether the hydrogen, oxygen and carbon dioxide molecules to be found in the atmosphere of the mars.

(d)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Calculate $20$ % of the escape velocity for mars.

$v=20 ⋅vescape100$ ........ (2)

Here, $v$ is the $20$ % of the escape velocity.

Calculation:

Substitute $5.0 km/s$ for $vescape$ in equation (2).

$v=20 ⋅( 5.0 km/s )100v= 1km/s$

Conclusion:

As $20$ % of the escape velocity is $1km/s$ , this is greater than $vrms$ for $O2$ and $CO2$ but less than for $H2$ . Thus $H2$ should not be present in the atmosphere of mars.

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Answer 2

Question

Chapter 17, Problem 26P

(a)

To determine

The rms speed of the $H2$ molecules.

(a)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$ ........ (1)

Here, $vrms$ is the root mean-square speed of the molecule, $R$ is the gas constant, $T$ is the temperature and $M$ is the molar mass.

Write the equation for the conversion of Celsius to kelvin.

$T(°K)=T(°C)+273.15$ ........ (2)

$T(°C)$ Is the temperature in Celsius and $T(°K)$ is the temperature in kelvin.

Calculation:

Substitute $0 °C$ for $T(°C)$ in equation (2).

$T(°K)=(0 °C)+273.15T(°K)=273.15 K$

Substitute $273.15 K$ for $T$ , $2×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K )( 273.15 K ) 2× 10 −3 kg/ mol vrms=1.84 Km/s$

Conclusion:

The rms speed of the $H2$ molecules is $1.84 Km/s$ .

(b)

To determine

The rms speed of $O2$ molecules.

(b)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$

Calculation:

Substitute $273.15 K$ for $T$ , $32×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K −1 )( 273.15 K ) 32× 10 −3 kg/ mol vrms= 461 m/s$

Conclusion:

The root mean square speed for the oxygen molecule is $461 m/s$ .

(c)

To determine

The root mean square speed of the $CO2$ molecule.

(c)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$

Calculation:

Substitute $273.15 K$ for $T$ , $44×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K −1 )( 273.15 K ) 44× 10 −3 kg/ mol vrms= 393 m/s$

Conclusion:

The root mean square speed for the $CO2$ molecule is $393 m/s$ .

(d)

To determine

Whether the hydrogen, oxygen and carbon dioxide molecules to be found in the atmosphere of the mars.

(d)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Calculate $20$ % of the escape velocity for mars.

$v=20 ⋅vescape100$ ........ (2)

Here, $v$ is the $20$ % of the escape velocity.

Calculation:

Substitute $5.0 km/s$ for $vescape$ in equation (2).

$v=20 ⋅( 5.0 km/s )100v= 1km/s$

Conclusion:

As $20$ % of the escape velocity is $1km/s$ , this is greater than $vrms$ for $O2$ and $CO2$ but less than for $H2$ . Thus $H2$ should not be present in the atmosphere of mars.

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Answer 3

Question

Chapter 17, Problem 26P

(a)

To determine

The rms speed of the $H2$ molecules.

(a)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$ ........ (1)

Here, $vrms$ is the root mean-square speed of the molecule, $R$ is the gas constant, $T$ is the temperature and $M$ is the molar mass.

Write the equation for the conversion of Celsius to kelvin.

$T(°K)=T(°C)+273.15$ ........ (2)

$T(°C)$ Is the temperature in Celsius and $T(°K)$ is the temperature in kelvin.

Calculation:

Substitute $0 °C$ for $T(°C)$ in equation (2).

$T(°K)=(0 °C)+273.15T(°K)=273.15 K$

Substitute $273.15 K$ for $T$ , $2×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K )( 273.15 K ) 2× 10 −3 kg/ mol vrms=1.84 Km/s$

Conclusion:

The rms speed of the $H2$ molecules is $1.84 Km/s$ .

(b)

To determine

The rms speed of $O2$ molecules.

(b)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$

Calculation:

Substitute $273.15 K$ for $T$ , $32×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K −1 )( 273.15 K ) 32× 10 −3 kg/ mol vrms= 461 m/s$

Conclusion:

The root mean square speed for the oxygen molecule is $461 m/s$ .

(c)

To determine

The root mean square speed of the $CO2$ molecule.

(c)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$

Calculation:

Substitute $273.15 K$ for $T$ , $44×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K −1 )( 273.15 K ) 44× 10 −3 kg/ mol vrms= 393 m/s$

Conclusion:

The root mean square speed for the $CO2$ molecule is $393 m/s$ .

(d)

To determine

Whether the hydrogen, oxygen and carbon dioxide molecules to be found in the atmosphere of the mars.

(d)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Calculate $20$ % of the escape velocity for mars.

$v=20 ⋅vescape100$ ........ (2)

Here, $v$ is the $20$ % of the escape velocity.

Calculation:

Substitute $5.0 km/s$ for $vescape$ in equation (2).

$v=20 ⋅( 5.0 km/s )100v= 1km/s$

Conclusion:

As $20$ % of the escape velocity is $1km/s$ , this is greater than $vrms$ for $O2$ and $CO2$ but less than for $H2$ . Thus $H2$ should not be present in the atmosphere of mars.

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Answer 4

Question

Chapter 17, Problem 26P

(a)

To determine

The rms speed of the $H2$ molecules.

(a)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$ ........ (1)

Here, $vrms$ is the root mean-square speed of the molecule, $R$ is the gas constant, $T$ is the temperature and $M$ is the molar mass.

Write the equation for the conversion of Celsius to kelvin.

$T(°K)=T(°C)+273.15$ ........ (2)

$T(°C)$ Is the temperature in Celsius and $T(°K)$ is the temperature in kelvin.

Calculation:

Substitute $0 °C$ for $T(°C)$ in equation (2).

$T(°K)=(0 °C)+273.15T(°K)=273.15 K$

Substitute $273.15 K$ for $T$ , $2×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K )( 273.15 K ) 2× 10 −3 kg/ mol vrms=1.84 Km/s$

Conclusion:

The rms speed of the $H2$ molecules is $1.84 Km/s$ .

(b)

To determine

The rms speed of $O2$ molecules.

(b)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$

Calculation:

Substitute $273.15 K$ for $T$ , $32×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K −1 )( 273.15 K ) 32× 10 −3 kg/ mol vrms= 461 m/s$

Conclusion:

The root mean square speed for the oxygen molecule is $461 m/s$ .

(c)

To determine

The root mean square speed of the $CO2$ molecule.

(c)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$

Calculation:

Substitute $273.15 K$ for $T$ , $44×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K −1 )( 273.15 K ) 44× 10 −3 kg/ mol vrms= 393 m/s$

Conclusion:

The root mean square speed for the $CO2$ molecule is $393 m/s$ .

(d)

To determine

Whether the hydrogen, oxygen and carbon dioxide molecules to be found in the atmosphere of the mars.

(d)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Calculate $20$ % of the escape velocity for mars.

$v=20 ⋅vescape100$ ........ (2)

Here, $v$ is the $20$ % of the escape velocity.

Calculation:

Substitute $5.0 km/s$ for $vescape$ in equation (2).

$v=20 ⋅( 5.0 km/s )100v= 1km/s$

Conclusion:

As $20$ % of the escape velocity is $1km/s$ , this is greater than $vrms$ for $O2$ and $CO2$ but less than for $H2$ . Thus $H2$ should not be present in the atmosphere of mars.

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Answer 5

Chapter 17, Problem 26P

(a)

To determine

The rms speed of the $H2$ molecules.

(a)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$ ........ (1)

Here, $vrms$ is the root mean-square speed of the molecule, $R$ is the gas constant, $T$ is the temperature and $M$ is the molar mass.

Write the equation for the conversion of Celsius to kelvin.

$T(°K)=T(°C)+273.15$ ........ (2)

$T(°C)$ Is the temperature in Celsius and $T(°K)$ is the temperature in kelvin.

Calculation:

Substitute $0 °C$ for $T(°C)$ in equation (2).

$T(°K)=(0 °C)+273.15T(°K)=273.15 K$

Substitute $273.15 K$ for $T$ , $2×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K )( 273.15 K ) 2× 10 −3 kg/ mol vrms=1.84 Km/s$

Conclusion:

The rms speed of the $H2$ molecules is $1.84 Km/s$ .

(b)

To determine

The rms speed of $O2$ molecules.

(b)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$

Calculation:

Substitute $273.15 K$ for $T$ , $32×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K −1 )( 273.15 K ) 32× 10 −3 kg/ mol vrms= 461 m/s$

Conclusion:

The root mean square speed for the oxygen molecule is $461 m/s$ .

(c)

To determine

The root mean square speed of the $CO2$ molecule.

(c)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$

Calculation:

Substitute $273.15 K$ for $T$ , $44×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K −1 )( 273.15 K ) 44× 10 −3 kg/ mol vrms= 393 m/s$

Conclusion:

The root mean square speed for the $CO2$ molecule is $393 m/s$ .

(d)

To determine

Whether the hydrogen, oxygen and carbon dioxide molecules to be found in the atmosphere of the mars.

(d)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Calculate $20$ % of the escape velocity for mars.

$v=20 ⋅vescape100$ ........ (2)

Here, $v$ is the $20$ % of the escape velocity.

Calculation:

Substitute $5.0 km/s$ for $vescape$ in equation (2).

$v=20 ⋅( 5.0 km/s )100v= 1km/s$

Conclusion:

As $20$ % of the escape velocity is $1km/s$ , this is greater than $vrms$ for $O2$ and $CO2$ but less than for $H2$ . Thus $H2$ should not be present in the atmosphere of mars.

Answer 6

Chapter 17, Problem 26P

(a)

To determine

The rms speed of the $H2$ molecules.

(a)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$ ........ (1)

Here, $vrms$ is the root mean-square speed of the molecule, $R$ is the gas constant, $T$ is the temperature and $M$ is the molar mass.

Write the equation for the conversion of Celsius to kelvin.

$T(°K)=T(°C)+273.15$ ........ (2)

$T(°C)$ Is the temperature in Celsius and $T(°K)$ is the temperature in kelvin.

Calculation:

Substitute $0 °C$ for $T(°C)$ in equation (2).

$T(°K)=(0 °C)+273.15T(°K)=273.15 K$

Substitute $273.15 K$ for $T$ , $2×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K )( 273.15 K ) 2× 10 −3 kg/ mol vrms=1.84 Km/s$

Conclusion:

The rms speed of the $H2$ molecules is $1.84 Km/s$ .

Answer 7

Chapter 17, Problem 26P

(a)

To determine

The rms speed of the $H2$ molecules.

Answer 8

(a)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$ ........ (1)

Here, $vrms$ is the root mean-square speed of the molecule, $R$ is the gas constant, $T$ is the temperature and $M$ is the molar mass.

Write the equation for the conversion of Celsius to kelvin.

$T(°K)=T(°C)+273.15$ ........ (2)

$T(°C)$ Is the temperature in Celsius and $T(°K)$ is the temperature in kelvin.

Calculation:

Substitute $0 °C$ for $T(°C)$ in equation (2).

$T(°K)=(0 °C)+273.15T(°K)=273.15 K$

Substitute $273.15 K$ for $T$ , $2×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K )( 273.15 K ) 2× 10 −3 kg/ mol vrms=1.84 Km/s$

Conclusion:

The rms speed of the $H2$ molecules is $1.84 Km/s$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

The rms speed of $O2$ molecules.

(b)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$

Calculation:

Substitute $273.15 K$ for $T$ , $32×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K −1 )( 273.15 K ) 32× 10 −3 kg/ mol vrms= 461 m/s$

Conclusion:

The root mean square speed for the oxygen molecule is $461 m/s$ .

Answer 13

(b)

To determine

The rms speed of $O2$ molecules.

Answer 14

(b)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$

Calculation:

Substitute $273.15 K$ for $T$ , $32×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K −1 )( 273.15 K ) 32× 10 −3 kg/ mol vrms= 461 m/s$

Conclusion:

The root mean square speed for the oxygen molecule is $461 m/s$ .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

To determine

The root mean square speed of the $CO2$ molecule.

(c)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$

Calculation:

Substitute $273.15 K$ for $T$ , $44×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K −1 )( 273.15 K ) 44× 10 −3 kg/ mol vrms= 393 m/s$

Conclusion:

The root mean square speed for the $CO2$ molecule is $393 m/s$ .

Answer 19

(c)

To determine

The root mean square speed of the $CO2$ molecule.

Answer 20

(c)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Write the expression for the rms speed of the molecule.

$vrms=3RTM$

Calculation:

Substitute $273.15 K$ for $T$ , $44×10−3 kg/mol$ for $M$ and $8.31 J/mol⋅K$ for $R$ in equation (1).

$vrms= 3( 8.31 Jmol -1 K −1 )( 273.15 K ) 44× 10 −3 kg/ mol vrms= 393 m/s$

Conclusion:

The root mean square speed for the $CO2$ molecule is $393 m/s$ .

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

To determine

Whether the hydrogen, oxygen and carbon dioxide molecules to be found in the atmosphere of the mars.

(d)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Calculate $20$ % of the escape velocity for mars.

$v=20 ⋅vescape100$ ........ (2)

Here, $v$ is the $20$ % of the escape velocity.

Calculation:

Substitute $5.0 km/s$ for $vescape$ in equation (2).

$v=20 ⋅( 5.0 km/s )100v= 1km/s$

Conclusion:

As $20$ % of the escape velocity is $1km/s$ , this is greater than $vrms$ for $O2$ and $CO2$ but less than for $H2$ . Thus $H2$ should not be present in the atmosphere of mars.

Answer 25

(d)

To determine

Whether the hydrogen, oxygen and carbon dioxide molecules to be found in the atmosphere of the mars.

Answer 26

(d)

Expert Solution

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0 km/s$ .

Temperature is $0 °C$ .

Formula used:

Calculate $20$ % of the escape velocity for mars.

$v=20 ⋅vescape100$ ........ (2)

Here, $v$ is the $20$ % of the escape velocity.

Calculation:

Substitute $5.0 km/s$ for $vescape$ in equation (2).

$v=20 ⋅( 5.0 km/s )100v= 1km/s$

Conclusion:

As $20$ % of the escape velocity is $1km/s$ , this is greater than $vrms$ for $O2$ and $CO2$ but less than for $H2$ . Thus $H2$ should not be present in the atmosphere of mars.