Problem 33 - Chemistry The Central Science

Question

Complete the following table for an ideal gas: $$\begin{array}{cccc}{P} & {v} & {n} & {T} \\ {2.00 \text { atm }} & {1.00 \mathrm{L}} & {0.500 \mathrm{mol}} & {\text { ?K }} \\ {0.300 \mathrm{atm}} & {0.250 \mathrm{L}} & {? \mathrm{mol}}\\\\{650 \text { torr }} & {\text { ?L }} & {0.333 \mathrm{mol}} & {350 \mathrm{K}} \\ {\text { ? atm }} & {585 \mathrm{mL}} & {0.250 \mathrm{mol}} & {295 \mathrm{K}}\end{array}$$

Step-by-Step Solution

Answer

\( T_1 = 49.4 \mathrm{K}, \: n_2 \: \text{cannot be determined}, \: V_3 = 6.307\: \mathrm{L}, \: P_4 = 3.40 \: \mathrm{atm} \)

Step 1: Calculate the temperature in the first row

Apply the ideal gas law for the first row: \( P_1V_1 = n_1RT_1 \). Then solve for \(T_1\): \( T_1 = \frac{P_1V_1}{n_1R} \). Now plug in the given values: \( T_1 = \frac{2.00 \mathrm{atm} \cdot 1.00\mathrm{L}}{0.500\mathrm{mol} \cdot 0.0821\frac{\mathrm{L}\mathrm{atm}}{\mathrm{mol}\mathrm{K}}} \). Step 2: Calculate the moles in the second row

Step 2: Calculate the moles in the second row

Apply the ideal gas law for the second row: \(P_2V_2 = n_2RT_2 \). Then solve for \(n_2\): \( n_2 = \frac{P_2V_2}{RT_2} \). Now plug the given values: \( n_2 = \frac{0.300 \mathrm{atm} \cdot 0.250\mathrm{L}}{0.0821\frac{\mathrm{L}\mathrm{atm}}{\mathrm{mol}\mathrm{K}} \cdot ? \mathrm{K}} \). Note that we cannot solve for \(n_2\) because the temperature is missing. Step 3: Calculate the volume in the third row

Step 3: Calculate the volume in the third row

Convert the pressure to atmospheres: \( P_3 = \frac{650 \mathrm{torr}}{760 \frac{\mathrm{torr}}{\mathrm{atm}}} \). Apply the ideal gas law for the third row: \(P_3V_3 = n_3RT_3 \). Then solve for \(V_3\): \( V_3 = \frac{n_3RT_3}{P_3} \). Now plug in the values: \( V_3 = \frac{0.333\mathrm{mol} \cdot 0.0821\frac{\mathrm{L}\mathrm{atm}}{\mathrm{mol}\mathrm{K}} \cdot 350\mathrm{K}}{\frac{650 \mathrm{torr}}{760 \frac{\mathrm{torr}}{\mathrm{atm}}}} \). Step 4: Calculate the pressure in the fourth row

Step 4: Calculate the pressure in the fourth row

Convert the volume to liters: \(V_4 = \frac{585 \mathrm{mL}}{1000 \frac{\mathrm{mL}}{\mathrm{L}}} \). Apply the ideal gas law for the fourth row: \(P_4V_4 = n_4RT_4 \). Then solve for \(P_4\): \( P_4 = \frac{n_4RT_4}{V_4} \). Now plug in the values: \( P_4 = \frac{0.250\mathrm{mol} \cdot 0.0821\frac{\mathrm{L}\mathrm{atm}}{\mathrm{mol}\mathrm{K}} \cdot 295\mathrm{K}}{\frac{585 \mathrm{mL}}{1000 \frac{\mathrm{mL}}{\mathrm{L}}}} \).

Key Concepts

Gas Laws

The Ideal Gas Law is a fundamental concept in chemistry and physics that relates four key properties of gases: pressure (P), volume (V), temperature (T), and the amount of gas in moles (n). This relationship is mathematically expressed as \( PV = nRT \), where \( R \) is the ideal gas constant. The Ideal Gas Law provides a simple formula to understand how these variables interact and allows for the calculation of any one property if the others are known.

Understanding this law is essential in solving practical problems involving gases, as it describes the behavior of an ideal gas—a theoretical gas composed of randomly moving point particles. While real gases can deviate from this behavior, especially at high pressures or low temperatures, the Ideal Gas Law is a good approximation under many conditions.

• Pressure is usually measured in atmospheres (atm) or torr.
• Volume is often measured in liters (L).
• Temperature must be in Kelvin (K) to use the Ideal Gas Law.
• The amount of gas is measured in moles.

Temperature Calculations

Temperature plays a crucial role in the Ideal Gas Law as it needs to be measured in an absolute scale, specifically Kelvin. This is because the Kelvin scale starts at absolute zero, the point where all molecular motion ceases. To convert Celsius to Kelvin, a simple addition is required: \( K = °C + 273.15 \).

In gas law calculations, especially for solving for temperature such as in the given problem's first row, knowing how to rearrange the Ideal Gas Law is key. By isolating temperature, we use the equation \( T = \frac{PV}{nR} \) to find the temperature, as was demonstrated in the solution. Keeping all variables in their proper units ensures accuracy.

• Always convert pressure to atm if needed.
• Ensure volume is in liters.
• Keep the mole value in moles.

Converting your values correctly will ensure your temperature calculations are accurate and meaningful.

Pressure Conversions

Pressure is another critical element in working with gases. It is important to convert between different units of pressure when needed. The two common units you may encounter are atmospheres (atm) and torr, where \(1 \, \text{atm} = 760 \, \text{torr}\).

In gas law problems, like converting the pressure from torr to atm seen in Step 3, using the conversion formula \( \text{P (atm)} = \frac{\text{P (torr)}}{760} \) ensures accuracy and uniformity in calculations. This conversion is necessary before plugging values into the Ideal Gas Law equation.

This procedure ensures that all units are consistent, which simplifies calculations and reduces errors. Keep track of your conversions!

• Set your pressure in atm before using it in the formula.
• Be mindful of unit conversions to keep consistency.

Mole Calculations

Moles in chemistry provide a measure of the amount of substance. In the context of gases, moles allow us to determine other properties like pressure, volume, or temperature. In the Ideal Gas Law, rearranging allows for calculations of moles if other values are known: \( n = \frac{PV}{RT} \).

In the problem's second row, solving for moles becomes challenging without an initial temperature. However, once temperature is obtained or if it is initially provided, substitutions into the rearranged formula provide much-needed values. Chemical reactions often involve quantities of gases, and calculating moles is critical to these types of problems.

Keep in mind that:

• Always use the same units: liters, atm, and Kelvin.
• Adjust the Ideal Gas Law properly to solve for moles.
• Remember that moles provide insight into amount and proportion in reactions.