Question

Given each of the following sets of values for three of the gas variables, calculate the unknown quantity. a. \(P=21.2\) atm; \(V=142 \mathrm{mL} ; n=0.432\) mol; \(T=\) \(? \mathrm{K}\) b. \(P=?\) atm; \(V=1.23 \mathrm{mL} ; n=0.000115\) mol; \(T=\) \(293 \mathrm{K}\) c. \(P=755 \mathrm{mm}\) Hg; \(V=? \mathrm{mL} ; n=0.473 \mathrm{mol} ; T=\) \(131^{\circ} \mathrm{C}\)

Short answer

The short answers to the given problems are as follows: a. The unknown temperature is approximately 92.96 K. b. The unknown pressure is approximately 2.75 atm. c. The unknown volume is approximately 16.24 L.

Step-by-step solution

Identify given values and the unknown variable

Here we are given: \(P = 21.2\) atm, \(V = 142\) mL, \(n = 0.432\) mol, and we need to find \(T\).

Convert the volume to liters

We need to convert the given volume from mL to L: \(1 L = 1000 mL\), so: \(V = \frac{142}{1000}L = 0.142 L\)

Isolate the unknown variable, T

Rearrange the ideal gas law equation to solve for the unknown temperature, \(T\): \(T = \frac{PV}{nR}\)

Substitute given values and solve for T

Substitute the given values and the gas constant into the equation to find \(T\): \(T = \frac{(21.2 \,\text{atm})(0.142\, \text{L})}{(0.432\, \text{mol})(0.08206\, \text{L atm K}^{-1}\text{mol}^{-1})} = 92.96K\) The unknown temperature is approximately 92.96 K. #b. Find the unknown pressure#

Identify given values and the unknown variable

Here we are given: \(V = 1.23\) mL, \(n = 0.000115\) mol, \(T = 293 K\), and we need to find \(P\).

Convert the volume to liters

We need to convert the given volume from mL to L as before: \(1 L = 1000 mL\), so: \(V = \frac{1.23}{1000}L = 0.00123 L\)

Isolate the unknown variable, P

Rearrange the ideal gas law equation to solve for the unknown pressure, \(P\): \(P = \frac{nRT}{V}\)

Substitute given values and solve for P

Substitute the given values and the gas constant into the equation to find \(P\): \(P = \frac{(0.000115\, \text{mol})(0.08206\, \text{L atm K}^{-1}\text{mol}^{-1})(293\, \text{K})}{0.00123\, \text{L}} = 2.75 \,\text{atm}\) The unknown pressure is approximately 2.75 atm. #c. Find the unknown volume#

Identify given values, the unknown variable, and convert temperature to Kelvin

Here we are given: \(P = 755\) mm Hg, \(n = 0.473\) mol, \(T = 131^{\circ} C\), and we need to find \(V\). Convert the temperature from Celsius to Kelvin: \[ T = 131^{\circ}\mathrm{C} + 273.15 = 404.15 \mathrm{K} \]

Convert pressure from mm Hg to atm

We need to convert the given pressure from mm Hg to atm: \(1 \,\text{atm} = 760 \,\text{mm Hg}\), so: \(P = \frac{755}{760}\, \text{atm} = 0.9934\, \text{atm}\)

Isolate the unknown variable, V

Rearrange the ideal gas law equation to solve for the unknown volume, \(V\): \(V = \frac{nRT}{P}\)

Substitute given values and solve for V

Substitute the given values and the gas constant into the equation to find \(V\): \(V = \frac{(0.473\, \text{mol})(0.08206\, \text{L atm K}^{-1}\text{mol}^{-1})(404.15\, \text{K})}{0.9934\, \text{atm}} = 16.24\, \text{L}\) The unknown volume is approximately 16.24 L.

Key concepts

Gas Variables

When dealing with gases, there are four primary variables to understand: pressure \(P\), volume \(V\), temperature \(T\), and the amount of gas present, usually measured in moles \(n\). Each of these variables plays a crucial role in determining the behavior of gas under different conditions.

• Pressure \(P\): It is the force exerted by the gas per unit area, commonly measured in atmospheres (atm), millimeters of mercury (mm Hg), or pascals (Pa).
• Volume \(V\): This measures the space the gas occupies, typically in liters (L) or milliliters (mL).
• Temperature \(T\): It's a measure of the average kinetic energy of gas particles, usually recorded in degrees Celsius (°C) or Kelvin (K).
• Moles \(n\): These indicate the quantity of gas molecules present in a sample.

These variables interrelate through the ideal gas law, expressed as \(PV = nRT\), where \(R\) is the ideal gas constant. Understanding how to manipulate this equation can help in solving for any of these unknown variables.

Pressure Calculation

Pressure calculation is often necessary for understanding how gases interact. The ideal gas law \(PV = nRT\) can be rearranged to solve for pressure as \(P = \frac{nRT}{V}\). This formula can help calculate the pressure exerted by a gas when the volume \(V\), amount \(n\), and temperature \(T\) are known.

• For example, given \(n = 0.000115\) mol, \(T = 293\, \text{K}\), and \(V = 0.00123\, \text{L}\), we substitute these values along with the gas constant \(R = 0.08206\, \text{L atm K}^{-1}\text{mol}^{-1}\) into the equation.
• Calculations show that \(P = \frac{(0.000115\, \text{mol})(0.08206\, \text{L atm K}^{-1}\text{mol}^{-1})(293\, \text{K})}{0.00123\, \text{L}} \approx 2.75 \,\text{atm}\).

It's also vital to convert units appropriately, such as mm Hg to atm, using the conversion \(1 \,\text{atm} = 760 \,\text{mm Hg}\) for consistency in calculations.

Volume Conversion

Volume conversion is a necessary step often encountered in gas calculations. Most gas laws use volume in liters (L), but in practice, volume might be provided in milliliters (mL).

• The conversion is straightforward: \(1\, \text{L} = 1000\, \text{mL}\). Therefore, to convert milliliters to liters, you divide the milliliters measurement by 1000.
• For instance, if you have a volume of \(142\, \text{mL}\), conversion to liters would be \(V = \frac{142}{1000} = 0.142\, \text{L}\).

Performing this conversion ensures that the volume is compatible with the gas constant \(R\) in calculations, maintaining consistency and accuracy.

Temperature Conversion

Temperature conversion is essential since the ideal gas law requires temperature in Kelvin (K), even though it might initially be given in Celsius (°C). The conversion is necessary for accurate calculations.

• To convert from Celsius to Kelvin, simply use the formula: \(T(\text{K}) = T(°\text{C}) + 273.15\).
• For example, if a temperature is provided as \(131^{\circ} \text{C}\), convert it by adding 273.15, resulting in \(T = 131 + 273.15 = 404.15\, \text{K}\).

Using Kelvin ensures linearity and proportionality in calculations, which are critical for the ideal gas law application.