Question

10 m3 of carbon dioxide is originally at a temperature of 50 °C and pressure of 10 kPa. Determine the new density and volume of the carbon dioxide if the temperature and pressure change to 75 oC and 15 kPa.

Answer 1

Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and
`7.2m^3` respectively.

Explanation :

First we have to calculate the new or final volume of carbon dioxide gas.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

where,

`P_1` = initial pressure of gas = 10 kPa

`P_2` = final pressure of gas = 15 kPa

`V_1` = initial volume of gas =
`10m^3`

`V_2` = final volume of gas = ?

`T_1` = initial temperature of gas =
`50^oC=273+50=323K`

`T_2` = final temperature of gas =
`75^oC=273+75=348K`

Now put all the given values in the above equation, we get:

`(10kPa* 10m^3)/(323K)=(15kPa* V_2)/(348K)`

`V_2=7.2m^3`

The new volume of carbon dioxide gas is
`7.2m^3`

Now we have to calculate the new density of carbon dioxide gas.

`PV=nRT\\\\PV=(m)/(M)RT\\\\P=(m)/(V)(RT)/(M)\\\\P=\rho (RT)/(M)\\\\\rho=(PM)/(RT)`

Formula for new density will be:

`\rho_2=(P_2M)/(RT_2)`

where,

`P_2` = new pressure of gas = 15 kPa

`T_2` = new temperature of gas =
`75^oC=273+75=348K`

M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

`\rho` = new density

Now put all the given values in the above equation, we get:

`\rho_2=((15kPa)* (44g/mole))/((8.314L.kPa/mol.K)* (348K))`

`\rho_2=0.2281g/L`

The new density of carbon dioxide gas is 0.2281 g/L