Question

A quantity of ideal gas at \(12.0^{\circ} \mathrm{C}\) and a pressure of \(108 \mathrm{kPa}\) occupies a volume of \(2.47 \mathrm{~m}^{3}\). (a) How many moles of the gas are present? \((b)\) If the pressure is now raised to \(316 \mathrm{kPa}\) and the temperature is raised to \(31.0^{\circ} \mathrm{C}\), how much volume will the gas now occupy? Assume there are no leaks.

Short answer

There are 108 moles of gas present initially and the volume of gas after the changes is 0.879 m³.

Step-by-step solution

Convert temperatures to Kelvin and pressures to Pascal

First, convert the temperatures from degrees Celsius to Kelvin using the formula \(K = C + 273.15\) which gives \(T_{1} = 285.15 K\) and \(T_{2} = 304.15 K\). Then, convert the pressures from kPa to Pa (since 1 kPa = 1000 Pa) which gives \(P_{1} = 108000 Pa\) and \(P_{2} = 316000 Pa\).

Calculate the number of moles in the first scenario

Using the gas law equation \(PV = nRT\) and solve for \(n\), we get \(n_1 = \frac{P_1V_1}{RT_1}\). Substituting \(P_1 = 108000 Pa\), \(V_1 = 2.47 m^3\), \(R = 8.314 J/(mol*K)\), and \(T_1 = 285.15 K\), we calculate \(n_1 = 107.96\) moles.

Calculate the volume in the second scenario

Using the same gas law equation and solve for \(V\), we get \(V_2 = \frac{n_2RT_2}{P_2}\). Since there are no leaks, so \(n_2 = n_1\). Substituting \(n_2 = 107.96\) mol, \(R = 8.314 J/(mol*K)\), \(T_2 = 304.15 K\), and \(P_2 = 316000 Pa\), we calculate \(V_2 = 0.879\) m³.

Round off to the appropriate number of significant figures

Since the given values are to 3 significant figures, we should also maintain 3 significant figures in the answer. So, \(n = 108\) moles and \(V_2 = 0.879\) m³.

Key concepts

Temperature Conversion

When dealing with gas laws, temperature must always be in Kelvin. This is crucial because Kelvin is an absolute temperature scale based on absolute zero, where no molecular movement occurs. To convert from Celsius to Kelvin, use the formula:

• Kelvin (K) = Celsius (°C) + 273.15

So, in our problem, we converted:

• Initial temperature: 12°C to 285.15 K
• New temperature: 31°C to 304.15 K

Remember, Kelvin does not have a degree symbol (°). These conversions ensure that temperatures are correctly used in gas law equations.

Pressure Conversion

Pressure in gas law calculations must be in Pascal (Pa) for compatibility with the ideal gas constant (R), which is usually given in J/(mol*K). Since the problem provides pressure in kilopascal (kPa), you must convert it. Here's how:

• 1 kPa = 1000 Pa

Thus, we convert:

• Initial pressure: 108 kPa to 108,000 Pa
• New pressure: 316 kPa to 316,000 Pa

This conversion allows us to use the ideal gas law, making calculations seamless and accurate!

Moles Calculation

The number of moles, indicated as 'n' in the ideal gas law, represents the amount of substance. To find this, we use:

• Ideal Gas Law: \( PV = nRT \)

Solving for \( n \), we have:

• \( n = \frac{PV}{RT} \)

Given:

• Pressure (P): 108,000 Pa
• Volume (V): 2.47 m³
• Temperature (T): 285.15 K
• Gas constant (R): 8.314 J/(mol*K)

This calculation yields 107.96 moles, but rounded to three significant figures as 108 moles. This helps conform to precision required in scientific measurements.

Volume Calculation

Calculating the new volume of gas at changed conditions also uses the ideal gas law. The formula rearranges to find volume as:

• \( V = \frac{nRT}{P} \)

Under the new conditions:

• Moles (n): 107.96 mol
• Pressure (P): 316,000 Pa
• Temperature (T): 304.15 K
• Gas constant (R): 8.314 J/(mol*K)

Plugging these values in, we find the new volume to be 0.879 m³. Ensure the precision matches the significant figures of the provided data, which in this case is three. This detail ensures answers maintain scientific accuracy and clarity.