Question

Complete the following table for an ideal gas: $$ \begin{array}{llll} \hline \boldsymbol{P} & \boldsymbol{V} & \boldsymbol{n} & \boldsymbol{T} \\ \hline 2.00 \mathrm{~atm} & 1.00 \mathrm{~L} & 0.500 \mathrm{~mol} & ? \mathrm{~K} \\ 0.300 \mathrm{~atm} & 0.250 \mathrm{~L} & ? \mathrm{~mol} & 27^{\circ} \mathrm{C} \\ 650 \text { torr } & ? \mathrm{~L} & 0.333 \mathrm{~mol} & 350 \mathrm{~K} \\ ? \mathrm{~atm} & 585 \mathrm{~mL} & 0.250 \mathrm{~mol} & 295 \mathrm{~K} \\ \hline \end{array} $$

Short answer

The completed table is: $$ \begin{array}{llll} \hline \boldsymbol{P} & \boldsymbol{V} & \boldsymbol{n} & \boldsymbol{T} \\ \hline 2.00 \mathrm{~atm} & 1.00 \mathrm{~L} & 0.500 \mathrm{~mol} & 48.60 \mathrm{~K} \\ 0.300 \mathrm{~atm} & 0.250 \mathrm{~L} & 0.0305 \mathrm{~mol} & 300.15 \mathrm{~K} \\ 0.855 \text{ atm } & 10.14 \mathrm{~L} & 0.333 \mathrm{~mol} & 350 \mathrm{~K} \\ 8.24 \mathrm{~atm} & 0.585 \mathrm{~L} & 0.250 \mathrm{~mol} & 295 \mathrm{~K} \\ \hline \end{array} $$

Step-by-step solution

Convert units if needed

For each row, we'll check if the given values are in the correct units (atm, L, mol, K) and convert them if needed. Row 1: No conversion needed. Row 2: Temperature: \(T = T+273.15 = 27 + 273.15 = 300.15 K\) Row 3: Pressure: \(P = \frac{650 \text{ torr}}{760} = 0.855 \text{ atm}\) Row 4: Volume: \(V = \frac{585 \text{ mL}}{1000} = 0.585 \text{ L}\)

Apply the ideal gas law equation to each row

For each row, we'll solve for the unknown variable using the given variables and the ideal gas law: \(PV = nRT\).

Row 1: Find temperature T

We have \(P = 2.00 \text{ atm}\), \(V = 1.00 \text{ L}\), \(n = 0.500 \text{ mol}\), and \(R = 0.0821 \frac{L \cdot atm}{mol \cdot K}\). We'll solve for T: \(T = \frac{PV}{nR} = \frac{(2.00)(1.00)}{(0.500)(0.0821)} = 48.60 K\)

Row 2: Find number of moles n

We have \(P = 0.300 \text{ atm}\), \(V = 0.250 \text{ L}\), \(T = 300.15 \text{ K}\), and \(R = 0.0821 \frac{L \cdot atm}{mol \cdot K}\). We'll solve for n: \(n = \frac{PV}{RT} = \frac{(0.300)(0.250)}{(0.0821)(300.15)} = 0.0305 \text{ mol}\)

Row 3: Find volume V

We have \(P = 0.855 \text{ atm}\), \(n = 0.333 \text{ mol}\), \(T = 350 \text{ K}\), and \(R = 0.0821 \frac{L \cdot atm}{mol \cdot K}\). We'll solve for V: \(V = \frac{nRT}{P} = \frac{(0.333)(0.0821)(350)}{(0.855)} = 10.14 \text{ L}\)

Row 4: Find pressure P

We have \(V = 0.585 \text{ L}\), \(n = 0.250 \text{ mol}\), \(T = 295 \text{ K}\), and \(R = 0.0821 \frac{L \cdot atm}{mol \cdot K}\). We'll solve for P: \(P = \frac{nRT}{V} = \frac{(0.250)(0.0821)(295)}{(0.585)} = 8.24 \text{ atm}\) So the completed table is : $$ \begin{array}{llll} \hline \boldsymbol{P} & \boldsymbol{V} & \boldsymbol{n} & \boldsymbol{T} \\ \hline 2.00 \mathrm{~atm} & 1.00 \mathrm{~L} & 0.500 \mathrm{~mol} & 48.60 \mathrm{~K} \\ 0.300 \mathrm{~atm} & 0.250 \mathrm{~L} & 0.0305 \mathrm{~mol} & 300.15 \mathrm{~K} \\ 0.855 \text{ atm } & 10.14 \mathrm{~L} & 0.333 \mathrm{~mol} & 350 \mathrm{~K} \\ 8.24 \mathrm{~atm} & 0.585 \mathrm{~L} & 0.250 \mathrm{~mol} & 295 \mathrm{~K} \\ \hline \end{array} $$

Key concepts

Gas Constant

In the realm of chemistry, the gas constant, often symbolized as \( R \), plays a crucial role. It serves as a bridge between various properties of gases. The ideal gas law equation, \( PV = nRT \), uses this constant to relate pressure \( P \), volume \( V \), the number of moles \( n \), and temperature \( T \).

The value of the gas constant \( R \) is typically \( 0.0821 \) L atm/mol K when working with pressure in atmospheres (atm), volume in liters (L), and temperature in Kelvin (K). This value allows scientists to derive one gas property when others are known. By using the correct units for \( R \), calculations become straightforward, especially when data is initially given in standard units (atm, L, mol, K).

Remember, choosing the appropriate value for \( R \) is critical because it impacts the outcome of your calculations. Keep an eye on unit consistency to ensure your results are accurate and meaningful.

Pressure Conversion

Often in gas law problems, pressure is not conveniently given in the desired unit. One of the most common conversions is between atmospheres (atm) and torr—where \( 1 \, atm \) equals \( 760 \, torr \). Such a conversion can be pivotal when setting up the ideal gas law equation properly.

For instance, if pressure is initially in torr, you must convert it to atm for consistency with the gas constant \( R \) in \( 0.0821 \) L atm/mol K. This conversion is done by dividing the pressure in torr by \( 760 \, (\text{e.g., } 650 \, \text{torr} = \frac{650}{760} \, atm) \).

By ensuring pressure is in the correct unit, calculations become more seamless. This approach helps prevent unnecessary errors and ensures rational and consistent results across different gas scenarios.

Volume Conversion

In scientific problems involving gases, volume conversions are often necessary to maintain consistency. When dealing with the ideal gas law, it's essential to express volume in liters (L) to match the units of the gas constant \( R \).

A common conversion is from milliliters (mL) to liters. Since \( 1000 \, mL = 1 \, L \), this is achieved by dividing the volume in mL by \( 1000 \) (for example, \( 585 \, mL = \frac{585}{1000} \, L = 0.585 \, L \)).

Ensuring that volume is in liters not only aligns with the units used in \( R \), but also facilitates direct application of the ideal gas law without further modifications. Clear unit conversion practice like this paves the way for accurate gas property evaluations.

Temperature Conversion

Temperature is a critical factor in the behavior of gases and must always be expressed in Kelvin for most gas calculations, including those using the ideal gas law. Kelvin is the standard SI unit for temperature in scientific equations, as it begins from absolute zero.

To convert from degrees Celsius (°C) to Kelvin (K), the formula is straightforward: add \( 273.15 \). For example, a temperature of \( 27^{\circ} \mathrm{C} \) converts to Kelvin as \( 27 + 273.15 = 300.15 \, K \).

This conversion ensures that all temperatures are positive, preventing any confusion or error in calculations due to negative temperatures. By standardizing temperature in Kelvin across all computations, you maintain clarity and precision when predicting gas behaviors.