Question

Which of the following choices represents the molar volume of an ideal gas at ${25}^{\circ }\mathrm{C}$ and 1.5 atm? (a) $(298\times 1.5/273)\times 22.4\mathrm{L};$ (b) $22.4\mathrm{L}$ (c) $(273\times 1.5/298)\times 22.4\mathrm{L}$ (d) $[298/(273\times 1.5)]\times 22.4\mathrm{L}$ (e) $[273/(298\times 1.5)]\times 22.4\mathrm{L}$

Short answer

The molar volume of an ideal gas at 25°C and 1.5 atm is approximately 20.7 L or, therefore, the closest choice is (c) $(273\ast 1.5/298)\times 22.4$ L.

Step-by-step solution

Convert Temperature to Kelvin

First, we'll need to convert the given temperature from Celsius to Kelvin. Add 273.15 to the Celsius temperature to do this: ${25}^{\circ }C+273.15=298.15K$. For simplicity, we will round this to 298 K.

Adjust Molar Volume

We know that at 1 atm and 273 K, the molar volume of an ideal gas is approximately 22.4 L. But here the temperature is 298 K and the pressure is 1.5 atm. This will affect the molar volume proportionally. It will increase with the temperature and decrease with the pressure. Therefore, we need to adjust the 22.4 L value to reflect the change in temperature and pressure. The correct formula for adjustment should be $(273\ast 1.5/298)\times 22.4$ L.

Calculate the New Molar Volume

Now plug in the known values to calculate the new molar volume under these conditions: $(273\ast 1.5/298)\times 22.4$ L = $20.7$ L. So, the molar volume of an ideal gas at 25°C and 1.5 atm is approximately $20.7$ L.

Key concepts

Molar Volume

Molar volume is an important concept when dealing with gases. It describes the volume that one mole of a substance occupies at a particular temperature and pressure. For an ideal gas, the molar volume at standard temperature and pressure (STP) — which is 0°C and 1 atm — is approximately 22.4 liters. In ideal conditions, which assume no interactions between gas molecules and that the molecules occupy no space themselves, this value is constant.
However, real-life conditions often differ. Temperature and pressure changes, like those in the problem where the temperature is 25°C (which is higher than STP) and the pressure is 1.5 atm (which is higher than STP), require adjustments to this ideal molar volume. The formula given in the problem, $273\times 1.5/298$ multiplied by 22.4 L, compensates for these non-standard conditions by modifying the standard molar volume accordingly.

Temperature Conversion

When working with gas laws like the Ideal Gas Law, it's often necessary to convert temperatures from degrees Celsius to Kelvin. This is because gas laws require absolute temperature scales. The Kelvin scale is an absolute temperature scale, meaning it starts at absolute zero, which is the point at which all molecular motion ceases. To convert Celsius to Kelvin, you simply add 273.15. For example, converting 25°C to Kelvin is straightforward: $25+273.15=298.15$ K. In exercises, like the one given, valuable simplification is done by rounding this to 298 K. This minor rounding aids in calculation simplicity without much impact on typical results. Always remember, working in Kelvin is crucial for accurate gas law calculations.

Pressure and Volume Relationship

The relationship between pressure and volume is a key concept derived from the Ideal Gas Law, which is $PV=nRT$. In this formula, $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal gas constant, and $T$ is temperature.For our example, when dealing with a gas at different pressures and temperatures, adjusting the volume is crucial. If the temperature is raised while pressure is also increased, as in the exercise situation, the volume changes accordingly. Specifically:

• Increased temperature generally leads to increased volume as molecules spread out more.
• Increased pressure, however, tends to compress the gas, reducing volume.

Thus, the given formula in the exercise, $(273\times 1.5/298)\times 22.4$ L, expertly adjusts for both temperature and pressure changes, depicting how the volume of the ideal gas is redistributed under new conditions.