Question

A shipment from Earth to Mars contains a 60 gallon​ [gal] tank filled with an ideal gas. The molecular weight of the ideal gas is 15.9 grams per mole​ [g/mol]. NASA tells the astronauts on Mars that the tank and gas combined had an initial total mass of 5 kilograms​ [kg] and the pressure was 3 atmospheres​ [atm] before it left the Earth. During the​ trip, the temperature was kept constant at 20 degrees Celsius ​[°​C]. When it arrives on​ Mars, the astronauts check the sensors and discover the pressure is 2.2 atmospheres​ [atm] and the temperature is 20 degrees Celsius ​[°​C]. Based on this​ information, the astronauts determine that some of the gas leaked during shipment. Determine how much gas leaked from the tank in units of grams​ [g]. Gravity on Mars​ = 3.71 meters per second squared ​[m/s2​].

Answer 1

The ideal gas equation can find the amount of gas lost is 7 mol

Given parameters

• The initial and final pressures P1 = 3 atm and P2 = 2.2 atm

• The temperature T = 20ºC

• Container volume V = 60 gallon

To find

• The moles of gas lost

The intentional system of measurements (SI) is a system that determines which are the fundamental units, this allows to carry out calculations and exchange measurements in a uniform way and without errors, let's reduce the magnitudes to the SI system

        P₁ = 3 atm (1 10⁵ Pa / 1 atm) = 3 10⁵ Pa

        P₂ = 2.2 atm (1 10⁵ / 1 atm) = 2.2 10⁵ Pa

        PM = 15.9 g / mol (1 kg / 1000 g) = 15.9 10⁻⁻³ kg / mol

        V = 60 gallon (1 m³ / 264.172 gal) = 0.2271 m³

         T = 20 + 273.15 = 293.15 K

Ideal gases are gases that have no interactions between them, so they can be described by the equation

          PV = n R T

Where P is the pressure, V the volume, n the number of moles, R the ideal gas constant (R = 8.314 $\frac{J}{mol \ K}$) and T the temperature.

Let's look for the initial moles, that is, on Earth

          n = $\frac{PV}{RT}$

          n = $\frac{3 \ 10^5 \ 0.2271 }{ 8.314 \ 293.15}$

          n = 27.95 mol

Let's write the ideal gas equation for the two instants let's use subscript 1 for Earth and subscript 2 when the spacecraft is on Tuesday

            P₁ V = n₁ R T

             P₂ V = n₂ R T

            $\frac{P_1}{n_1} = \frac{P_2}{n_2}$

            n₂ = $\frac{P_2}{P_1} \ n_1$  

            n₂ = $\frac{2.2 \ 10^5 }{ 3 \ 10^5} \ 27.95$

            n₂ = 20.95 mol

This is the amount of moles left, the moles lost are

           n_ {lost} = n₁ -n₂

           n_ {lost} = 27.95 - 20.95

           n_ {lost} = 7 mol

Using the ideal gas equation we can find the amount of gas lost is 7 mol

Learn more about ideal gas here:

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