What is the pressure exerted by 4.5m\(^3\) of gas at 17ºC in a cylinder if the number of moles is 8.3 moles? (R = 8.31 JK\(^{-1}\)mol\(^{-1}\))

Answer Details

This is a direct application of the ideal gas equation in molar form:

\[PV = nRT \quad\Rightarrow\quad P = \frac{nRT}{V}.\]

The one conversion that must be made is the temperature, because \(T\) in this equation is the absolute temperature:

\[T = 17 + 273 = 290\ \text{K}.\]

Now substitute, keeping the units consistent in the SI system (\(V\) in \(\text{m}^3\), \(R\) in \(\text{J K}^{-1}\text{mol}^{-1}\), giving \(P\) in pascals):

\[P = \frac{8.3\times 8.31\times 290}{4.5} = \frac{20\,002}{4.5} = 4445\ \text{Pa}.\]

The pressure is about \(4445\ \text{Pa}\).

The commonest error is substituting \(17\) for the temperature, which gives roughly \(260\ \text{Pa}\), a value so small it should look wrong at once. A second slip is confusing the two very similar numbers in the data: \(8.3\) is the number of moles while \(8.31\ \text{J K}^{-1}\text{mol}^{-1}\) is the molar gas constant, and both appear in the numerator, so neither may be dropped. Before dividing, check that only the volume sits in the denominator, and always convert Celsius to kelvin as your first line of working in any gas calculation.