Question

A 0.0555 mol sample of Kr has a temperature of \(188^{\circ} \mathrm{C}\) and a volume of \(0.577 \mathrm{~L}\). What pressure does it have?

Short answer

The pressure is 3.64 atm.

Step-by-step solution

Convert Temperature to Kelvin

First, convert the temperature from Celsius to Kelvin since gas law calculations require the temperature in Kelvin.The formula to convert Celsius to Kelvin is:\[ T(K) = T(°C) + 273.15 \]Substitute the given temperature:\[ T(K) = 188 + 273.15 = 461.15 \, K \].

Write Down the Ideal Gas Law

Use the Ideal Gas Law to relate the pressure, volume, number of moles, and temperature of a gas.The Ideal Gas Law formula is:\[ PV = nRT \]where:- \(P\) is the pressure in atm,- \(V\) is the volume in liters,- \(n\) is the number of moles,- \(R\) is the ideal gas constant (0.0821 \(L \cdot atm \cdot mol^{-1} \cdot K^{-1}\)),- \(T\) is the temperature in Kelvin.

Rearrange the Ideal Gas Law to Solve for Pressure

Rearrange the ideal gas law formula to solve for pressure (\(P\)):\[ P = \frac{nRT}{V} \].

Substitute the Known Values

Substitute the values given in the problem into the rearranged equation:- \(n = 0.0555 \) mol,- \(R = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K^{-1}\),- \(T = 461.15\, K\),- \(V = 0.577\, L\).\[ P = \frac{0.0555 \times 0.0821 \times 461.15}{0.577} \].

Calculate the Pressure

Calculate the pressure using the values in the formula:\[ P = \frac{0.0555 \times 0.0821 \times 461.15}{0.577} = 3.64 \, atm \].

Key concepts

Temperature Conversion

When solving problems involving gases, the temperature must be in Kelvin, because Kelvin is an absolute temperature scale. Converting from Celsius to Kelvin is straightforward. Simply take the temperature in Celsius and add 273.15. This adjustment accounts for the absolute zero reference point in Kelvin, which is -273.15°C.
For example, if the temperature is given as 188°C, you would calculate \[ T(K) = 188 + 273.15 = 461.15 \, K \].
With this conversion, all gas law calculations will yield correct results.

Gas Constant

The ideal gas constant is a crucial factor in calculations involving gases. It appears in the Ideal Gas Law formula: \[ PV = nRT \].
The gas constant, denoted as \(R\), bridges the relationship between pressure, volume, number of moles, and temperature. It is essential to use consistent units throughout your calculations.

• In our example, \(R = 0.0821 \, L \cdot atm \cdot mol^{-1} \cdot K^{-1}\).
• This constant varies based on the units of pressure, printed as atm in this case, which matches our pressure unit.

Always make sure to match \(R\)'s units with those of your other variables.

Pressure Calculation

The primary objective is to find the pressure of the gas sample, using the Ideal Gas Law: \[ PV = nRT \].
To isolate pressure \(P\), rearrange the formula: \[ P = \frac{nRT}{V} \].
This structure makes pressure directly proportional to the number of moles and temperature, and inversely proportional to volume.

• With known values —\( n = 0.0555 \) mol, \( R = 0.0821 \) L\cdot\,atm\cdot\,mol\(^{-1}\)\cdot\,K\(^{-1}\), \( T = 461.15\, K\), and \( V = 0.577\, L\)— you're equipped to compute pressure.
• This relationship underpins the behavior of gases where increasing the number of moles or temperature tends to raise the pressure if volume remains unchanged.

Performing the actual calculation, we find: \[ P = \frac{0.0555 \times 0.0821 \times 461.15}{0.577} = 3.64 \, atm \].

Moles of Gas

Understanding the number of moles \(n\) is fundamental when working with gases. Moles quantify the amount of a substance and play a central role in the Ideal Gas Law.
For the given exercise, we have 0.0555 moles of krypton gas. This value is used in the Ideal Gas Law to determine the gas's pressure, given the other conditions.
In essence, knowing the number of moles helps establish how many molecules or atoms interact in a gas sample, affecting how those particles exert pressure and occupy volume under specified temperature conditions.
Accurate calculations depend on the moles connecting amounts of substance to their macroscopic properties as outlined in the equations.