Question

At what temperature is a gas if 0.0851 moles of it is contained in a 604 mLvessel at 100.4 kPa?

Answer 1

Assuming the gas in question behaves as an ideal gas, we can use the Ideal Gas Law:

`PV=nRT`

Where P is the pressure, V is the volume, n is the number of moles, R is the Ideal Gas Constant and T is the absolute temperature.

We have to match the units from R and the other variables.

The pressure is in kPa, so we can look for a R value that is also in kPa.

One such value is:

`R\approx8.31446(L\cdot kPa)/(K\cdot mol)`

But the volume unit is in L, so we need to convert the volume we have:

`V=604mL=0.604L`

Now, we have:

`\begin{gathered} P=100.4kPa \\ V=0.604L \\ n=0.0851mol \\ R\approx8.31446(L\cdot kPa)/(K\cdot mol) \end{gathered}`

Solving the equation for T and substituting the values, we have:

`\begin{gathered} PV=nRT \\ T=(PV)/(nR) \\ T=(100.4\cdot kPa\cdot0.604\cdot L)/(8.31446\cdot L\cdot kPa\cdot K^(-1)\cdot mol^(-1)\cdot0.0851mol) \\ T=\frac{100.4\cdot0.604}{8.31446^{}^{}\cdot0.0851}K \\ T=(60.6416)/(0.707560\ldots)K \\ T=85.70517\ldots K \\ T\approx85.7K \end{gathered}`

So, the temperature is approximately 85.7 K.