Question

The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -41 oC and 900 Pa, respectively. (a) Determine the density of the Martian atmosphere for these conditions if the gas constant for the Martian atmosphere is assumed to be equivalent to that of carbon dioxide. (b) Compare the answer from part (a) with the density of the earth’s atmosphere during a spring day when the temperature is 18 oC and the pressure 101.6 kPa (abs).

Answer 1

To determine the density of the Martian atmosphere, we can use the Ideal Gas Law. Use the given pressure, temperature, and the gas constant for carbon dioxide to calculate the number of moles of CO2. To compare the density of the Martian atmosphere with that of the Earth's atmosphere, use the same approach.

Step-by-step explanation:

To determine the density of the Martian atmosphere, we can use the Ideal Gas Law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

(a) We can use the given pressure, temperature, and the gas constant for carbon dioxide to calculate the number of moles of CO2. Then, we can divide the number of moles by the volume to calculate the density of the Martian atmosphere.

(b) To compare the density of the Martian atmosphere with that of the Earth's atmosphere, we can use the same approach. Use the given pressure, temperature, and the gas constant for air to calculate the density of the Earth's atmosphere.

Answer 2

Part a)

`\rho = 0.0205 kg/m^3`

Part b)

`\rho = 1.22 kg/m^3`

So density of atmosphere at Martian Surface is very less than the density at Earth.

Step-by-step explanation:

Part a)

As per ideal gas equation we know that

`PM = \rho RT`

here we know that Martian atmosphere is equivalent to that of carbon

so we will have

`P = 900 Pa`

`T = 273 - 41 = 232 K`

now we will have

`(900)(0.044) = \rho (8.31)(232)`

`\rho = 0.0205 kg/m^3`

Part b)

Now for the earth surface the density of air is given for

`P = 101.6 kPa`

`T = 18 ^oC`

so we will have

`PM = \rho RT`

`(101.6* 10^3)(0.029) = \rho(8.31)(273 + 18)`

`\rho = 1.22 kg/m^3`

So density of atmosphere at Martian Surface is very less than the density at Earth.