Question

Suppose you have 0.100 m3 of co2, at pressure of 2.00 atm, and a temperature of 30.0◦c. how many grams of co2 do you have? express your answer in grams.

Answer 1

Answer: 353.936 grams of
`CO_2` will be there.

Explanation: Using Ideal gas equation, which is:

`PV=nRT`

Given:

P = 2 atm

`V=0.1m^3=100L` (Conversion Factor:
`1m^3=1000L` )

`T=30^oC=(273+30)K=303K` (Conversion factor = 0°C = 273 K)

`R=0.082057\text{ L atm }mol^(-1)K^(-1)` (Gas Constant)

Putting all the values in above equation, we calculate the number of moles.

`2atm* 100L=n(0.082057\text{ L atm }mol^(-1)K^(-1))(303K)`

n = 8.044 moles

Now, to calculate the grams of
`CO_2`, we use the formula:

`Moles=\frac{\text{Given mass}}{\text{Molar mass}}`

Molar mass of
`CO_2=44g/mol`

Putting the values in above equation, we get

`8.044mol=\frac{\text{Given mass}}{44g/mol}`

Mass of
`CO_2` = 353.936 g

Answer 2

Use Ideal Gas equation since values are given in Pressure (P), Volume (V), Temperature (T). P V = n R T ; n= PV / RT
To choose the value of R ( gas constant), check out the units of other values in the equation. They are in atm,m³, kelvin, and mole. So, its value is 8.205 m³ atm K⁻¹ mol⁻¹

P= 2.00 atmV= 0.100 m³R= 8.205 m³ atm K⁻¹ mol⁻¹T= (30+ 273) K ( recall Kelvin = Celcius + 273; K= °C + 273)
Substitute values into equation of n= PV / RT

n=2.00 atm * 0.100 m³ / 8.205 m³ atm K⁻¹ mol⁻¹ * 303 K = 8.04 * 10⁻⁵ mole

Mole of CO₂ is 8.04 * 10⁻⁵. Its molar mass is 44 g/mol.

To find out its mass use mole= mass / Molar mass ; n= m/ M; m= n M

m= (8.04 * 10⁻⁵ mol) (44 g/mol) = 0. 035 g

The answer is 0.035 g