Answer to Question #162867 in General Chemistry for Gabby

Question

1. We talked about the composition of the air on Earth- so now, let's look at Mars! Mars has an atmosphere that is 96% CO2, 1.9% N2, 1.9% Ar, and 0.1% O2. 

If the first manned mission manages to collect a 1.00-liter jar of Martian air at the ambient pressure of 0.6 kPa and ambient temperature of -60ºC, how many moles of gas were collected?

Answer 1

Q162867

1. We talked about the composition of the air on Earth- so now, let's look at Mars! Mars has an atmosphere that is 96% CO₂, 1.9% N₂, 1.9% Ar, and 0.1% O₂. 

If the first manned mission manages to collect a 1.00-liter jar of Martian air at the ambient pressure of 0.6 kPa and ambient temperature of -60ºC, how many moles of gas were collected?

Solution:

We are given the volume of the jar in which Martian air is collected.

The pressure and temperature is also given to us.

We will use the ideal gas equation PV = nRT and find the moles of gas present in the

jar.

Ideal gas equation is , PV = n RT ;

where,

P is the pressure is 'atm'.

V is the volume in 'L'

R is the ideal gas constant and it is equal to 0.08206 L-atm/mol-K.

T is the temperature in Kelvin.

n is the moles of the gas.

P = 0.6 kPa

Convert it to 'atm' by using the conversion factor

1kPa = 1000 Pa and 1atm = 101325 Pascal

P in 'atm' $= 0.6kPa * \frac{1000Pa}{1kPa}* \frac{1atm }{101325Pa} = 0.0059215 atm.$

T in Kelvin = -60ºC + 273.15 = 213.15K.

V = 1.00L

R = 0.08206 L-atm/mol-K

plug all this information in the ideal gas equation, PV = n RT

$0.0059215 atm * 1.00L = n * 0.08206 \frac{L.atm}{mol-K} * 213.15K ;$

$0.0059215\space L.atm = n * 17.491 \frac{L.atm}{mol};$

divide both the side by 17.491 L.atm/mol we have

$\frac{0.0059215\space L.atm }{17.491L.atm/mol} = \frac{n*17.491L.atm/mol}{17.491L.atm/mol};$

n = 3.3855 * 10^-4 moles.

We will write our answer in 2 significant figure.

Hence the moles of gas collected is 3.4 * 10^-4 moles.