Question

Swift has a 5.55mol sample of carbon dioxide. If the sample is 24.29kpa, and the volume is 525.79mL, what will the temperature of the sample be in Kelvin?

Answer 1

The first step is to convert the kPa to atm, using a conversion factor:

`24.29kPa\cdot(1atm)/(101.325kPa)=0.24atm`

Also convert the ml to L, using a conversion factor:

`525.79ml\cdot(1l)/(1000ml)=0.52579l`

Use the ideal gas law to find the temperature of the gas:

`\begin{gathered} Pv=nRT \\ T=(Pv)/(nR) \end{gathered}`

Where T is the temperature, P is the pressure, v is the volume, n is the number of moles and R is the ideal gas constant (0.082atm*L/mol*K):

`T=(0.24atm\cdot0.52579l)/(5.55mol\cdot0.082(atm\cdot L)/(mol\cdot K))=0.28K`

The answer is 0.28K.