Question

At what temperature is the molar volume of an ideal gas equal to \(22.4 \mathrm{L},\) if the pressure of the gas is \(2.5 \mathrm{atm} ?\)

Short answer

The temperature at which the molar volume of an ideal gas is equal to \(22.4 L\) and the pressure is \(2.5 atm\) is approximately \(684.16 K\).

Step-by-step solution

Understand the data given in the problem

Firstly, understand what is given in the problem. The molar volume, which is the volume per mole, is given as \(22.4 L\). We also know the pressure of the gas is \(2.5 atm\). Given this is a molar volume, we know that n = 1 mole. We're asked to find the temperature at which these conditions are met.

Substitution into the Ideal Gas Law formula

Next, substitute these values into the Ideal Gas Law formula, \(PV = nRT\). This means that \(2.5 atm * 22.4L = 1 mole * R * T\). We need to find the temperature T. The value of the ideal gas constant R depends on the units used; in this case, \(0.0821 L.atm/(mol.K)\) should be used.

Calculate the temperature

For the final step, rearrange the equation to solve for T: \(T = PV/(nR) = (2.5 atm * 22.4L) / (1 mol * 0.0821 L.atm/(mol.K))\). After performing the calculation, the result is approximately \(684.16 K\).

Key concepts

Molar Volume

The molar volume of a gas is the volume occupied by one mole of the gas at a given set of conditions, typically measured in liters. At standard temperature and pressure (STP), which is 0°C (273.15 K) and 1 atm pressure, the molar volume is generally taken to be 22.4 L for an ideal gas. This means that under these conditions, one mole of any ideal gas will occupy a volume of 22.4 liters.

In the exercise, the molar volume is specified as 22.4 L, but the pressure is not at STP. Instead, the given pressure is 2.5 atm. The concept of molar volume helps us set the scenario for using the ideal gas law because it simplifies our calculations by letting us assume the presence of exactly one mole of gas. Understanding molar volume is crucial because it acts as a starting point in many chemistry problems that involve gases.

Ideal Gas Constant

The ideal gas constant, often denoted as \( R \), is a crucial component of the ideal gas law equation, \( PV = nRT \). This constant helps in linking the units of pressure, volume, temperature, and amount of gas.

There are several versions of \( R \), determined by the units used. For an equation involving liters, atmospheres, moles, and Kelvin, \( R \) usually takes the value of \( 0.0821 \ \text{L.atm/(mol.K)} \).

This value ensures that the units cancel out properly, leaving us with temperature in Kelvin when solving the problem. The ideal gas constant is central to converting the physical conditions such as pressure and volume into useful thermal data, measurable in absolute temperature. In the exercise provided, using \( 0.0821 \ \text{L.atm/(mol.K)} \) precisely enables us to compute the correct temperature required for the given conditions.

Temperature Calculation

In problems involving the ideal gas law, calculating the temperature requires some careful rearrangement of the formula. Here, you start with the law \( PV = nRT \), where pressure \( P \), volume \( V \), number of moles \( n \), and the ideal gas constant \( R \) are known variables. The task is to solve for the temperature \( T \).

To isolate \( T \), rearrange the equation to \( T = \frac{PV}{nR} \). By substituting the values given:

• \( P = 2.5 \ \text{atm} \)
• \( V = 22.4 \ \text{L} \)
• \( n = 1 \ \text{mol} \)

and \( R = 0.0821 \ \text{L.atm/(mol.K)} \) into the equation, you can solve for \( T \).

Performing these calculations, \( T = \frac{2.5 \cdot 22.4}{1 \cdot 0.0821} \approx 684.16 \ \text{K} \). This calculation translates the physical state of the gas into temperature, rewarding you with a deeper understanding of how gases behave under varied conditions.