What's the pressure of an ideal gas if 3.5 mol occupy \(2.0 \mathrm{L}\) at \(-150^{\circ} \mathrm{C} ?\)

Measuring temperature for gas behavior requires consistency, which is why the Kelvin scale is the standard for scientific calculations. To convert a temperature from Celsius to Kelvin, which is a critical first step when working with the ideal gas law, you need to add 273.15 to the Celsius value. This adjustment is because the Kelvin scale starts at absolute zero, the point where molecular motion ceases.

For instance, if you have a temperature of \( -150^\circ C \), the conversion to Kelvin would be \( -150^\circ C + 273.15 = 123.15 K \). By using Kelvin, we ensure that our temperature measurements reflect the absolute scale that correlates directly with the energy of the gas particles.

The ideal gas law, expressed as \( PV = nRT \), describes the relationship between pressure (\(P\)), volume (\(V\)), number of moles of gas (\(n\)), temperature in Kelvin (\(T\)), and the ideal gas constant (\(R\)). It's crucial to identify what values are provided in an exercise to apply this law accurately.

Let's visualize with a practical example. If told that 3.5 moles of a gas occupy 2.0 liters at -150 degrees Celsius, you'll first have to convert the temperature as was done in the previous section. You now have the variables: \(n = 3.5 \, mol\), \(V = 2.0 \, L\), and \(T = 123.15 \, K\). The ideal gas constant \(R\) value depends on the units of pressure and volume being used and is commonly \(0.0821 \, L \cdot atm / (K \cdot mol)\) for liters and atmospheres.

Once you've pinpointed your variables, the next step in solving an ideal gas law problem is the substitution of these values into the ideal gas law equation. Following the previously identified values, the substitution process is fairly straightforward. You replace the variables with the corresponding figures.

For the given problem, we substitute into the equation as follows: \(P = \frac{nRT}{V}\). Plugging in the values, we get \(P = \frac{(3.5 \, mol) \cdot (0.0821 \, L \cdot atm / (K \cdot mol)) \cdot (123.15 \, K)}{2.0 \, L}\). You can then compute the final pressure in atmospheres through simple arithmetic. Detailed comprehension of each step and careful substitution help to prevent errors and ensure a correct solution.