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

Expert verified
The completed table for the ideal gas is as follows: $$ \begin{array}{llll} \hline \boldsymbol{P} & \boldsymbol{v} & \boldsymbol{n} & \boldsymbol{T} \\ \hline 2.00 \mathrm{~atm} & 1.00 \mathrm{~L} & 0.500 \mathrm{~mol} & 48.66\mathrm{~K}\\ 0.300 \mathrm{~atm} & 0.250 \mathrm{~L} & 0.0031 \mathrm{~mol} & 300\mathrm{~K} \\ 650 \mathrm{~torr} & 10.18 \mathrm{~L} & 0.333 \mathrm{~mol}& 350\mathrm{~K}\\ 10.49 \mathrm{~atm} & 0.585 \mathrm{~L} & 0.250 \mathrm{~mol} & 295\mathrm{~K} \\ \hline \end{array} $$

Step by step solution

01

Case 1: Missing Temperature (T)

We have P = 2.00 atm, V = 1.00 L, n = 0.500 mol, and we need to find T. Rearrange the Ideal Gas Law equation to solve for T: \(T = \frac{PV}{nR}\) Now, plug in the given values: \(T = \frac{(2.00 \, atm)(1.00 \, L)}{(0.500 \, mol)(0.0821 \, L \, atm/mol \, K)} = 48.66 \, K\)
02

Case 2: Missing Number of Moles (n)

We have P = 0.300 atm, V = 0.250 L, T = 27°C (= 300 K, by adding 273 to convert to Kelvin), and we need to find n. Rearrange the Ideal Gas Law equation to solve for n: \(n = \frac{PV}{RT}\) Now, plug in the given values: \(n = \frac{(0.300 \, atm)(0.250 \, L)}{(0.0821 \, L \, atm/mol \, K)(300 \, K)} = 0.0031 \, mol\)
03

Case 3: Missing Volume (V)

We have P = 650 torr, n = 0.333 mol, T = 350 K, and we need to find V. First, convert the pressure P from torr to atm: \(P = \frac{650 \, torr}{760 \, torr/atm} = 0.8553 \, atm\) Now, rearrange the Ideal Gas Law equation to solve for V: \(V = \frac{nRT}{P}\) Plug in the given values: \(V = \frac{(0.333 \, mol)(0.0821 \, L \, atm/mol \, K)(350 \, K)}{(0.8553 \, atm)} = 10.18 \, L\)
04

Case 4: Missing Pressure (P)

We have V = 585 mL, n = 0.250 mol, T = 295 K, and we need to find P. First, convert the volume V from mL to L: \(V = \frac{585 \, mL}{1000 \, mL/L} = 0.585 \, L\) Now, rearrange the Ideal Gas Law equation to solve for P: \(P = \frac{nRT}{V}\) Plug in the given values: \(P = \frac{(0.250 \, mol)(0.0821 \, L \, atm/mol \, K)(295 \, K)}{(0.585 \, L)} = 10.49 \, atm\) Here's the completed table: $$ \begin{array}{llll} \hline \boldsymbol{P} & \boldsymbol{v} & \boldsymbol{n} & \boldsymbol{T} \\ \hline 2.00 \mathrm{~atm} & 1.00 \mathrm{~L} & 0.500 \mathrm{~mol} & 48.66\mathrm{~K}\\ 0.300 \mathrm{~atm} & 0.250 \mathrm{~L} & 0.0031 \mathrm{~mol} & 300\mathrm{~K} \\ 650 \mathrm{~torr} & 10.18 \mathrm{~L} & 0.333 \mathrm{~mol}& 350\mathrm{~K}\\ 10.49 \mathrm{~atm} & 0.585 \mathrm{~L} & 0.250 \mathrm{~mol} & 295\mathrm{~K} \\ \hline \end{array} $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Gas Pressure
Gas pressure is a critical concept in the study of gases and comes up frequently when dealing with the Ideal Gas Law. At its simplest, pressure is defined as the force exerted by a gas per unit area of the container it’s in. This force arises as countless gas molecules collide with the walls of their container. These collisions are the root cause of the pressure we measure, such as when we fill up a tire or check the air in a balloon.

When working with the Ideal Gas Law, which is \( PV=nRT \), pressure (\(P\)) will typically be measured in atmospheres (atm) or torr. One atmosphere is equivalent to the average pressure at sea level and is a standard unit for measuring gas pressure. However, if the pressure is given in torr, it's crucial to convert it to atmospheres to maintain consistency in the units used in the equation (1 atm = 760 torr). An understanding of how to manipulate and measure gas pressure is essential when predicting the behavior of gases under different conditions.
Molar Volume and Its Significance
Molar volume, denoted by \(v\), is the volume occupied by one mole of a substance, and in the case of gases, we typically refer to it under standard conditions of temperature and pressure. For ideal gases, the molar volume at standard temperature and pressure (0°C and 1 atm) is about 22.4 liters.

However, molar volume can change with varying conditions and understanding this is pivotal in using the Ideal Gas Law effectively. The Ideal Gas Law provides an equation that allows us to calculate the volume a gas occupies, taking into account the number of moles, temperature, and pressure. The direct relationship between volume and both temperature (\(T\)) and number of moles (\(n\)), and its inverse relationship with pressure, allows us to predict how the volume of a gas will change in different scenarios. Thus, a clear grasp of the concept of molar volume enhances one's ability to interpret and predict gas behaviors accurately.
The Role of Gas Temperature in the Ideal Gas Law
Temperature (typically denoted as \(T\)) is a measure of the average kinetic energy of the molecules in a substance, and in the context of gases, it’s a key factor in determining their behavior. For the purposes of calculations using the Ideal Gas Law, temperature must always be in Kelvin. This unit of measurement is chosen because it's an absolute scale which means 0 Kelvin (−273.15°C) is the point where the particles of an ideal gas would have no kinetic energy, essentially a theoretical point where all motion stops.

To convert Celsius to Kelvin, one must add 273 to the Celsius temperature. This transformation is crucial because if the temperature is incorrectly used in Celsius when computing the Ideal Gas Law, the results will be false, leading to a misunderstanding of gas behavior. A correct appreciation of gas temperature is important for predicting how a gas would expand or compress under different thermal conditions.
Number of Moles: Quantifying Gas Amounts
The number of moles (\(n\)) is a unit of measure that quantifies the amount of a substance. In terms of gases, a mole is Avogadro's number (approximately \(6.022 \times 10^{23}\)) of molecules. The number of moles is a critical variable in the Ideal Gas Law as it provides a direct measure of the quantity of gas involved.

The Ideal Gas Law equation reveals that the amount of gas (moles) is directly proportional to the volume (\(V\)) and temperature (\(T\)), while inversely proportional to the pressure (\(P\)). Understanding the interplay between the number of moles and these conditions can give us insights into gas density, as well as how to calculate the needed amount of gas for a reaction or predict the outcomes of mixing different volumes of gases at various pressures and temperatures. Being able to calculate and interpret the number of moles is key for anyone studying chemistry or working with gases in a scientific context.

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Most popular questions from this chapter

Hydrogen gas is produced when zinc reacts with sulfuric acid: $$ \mathrm{Zn}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{H}_{2}(g) $$ If \(159 \mathrm{~mL}\) of wet \(\mathrm{H}_{2}\) is collected over water at \(24{ }^{\circ} \mathrm{C}\) and a barometric pressure of 738 torr, how many grams of \(Z\) n have been consumed? (The vapor pressure of water is tabulated in Appendix B.)

A glass vessel fitted with a stopcock valve has a mass of \(337.428 \mathrm{~g}\) when evacuated. When filled with \(\mathrm{Ar}\), it has a mass of \(339.854 \mathrm{~g}\). When evacuated and refilled with a mixture of Ne and Ar, under the same conditions of temperature and pressure, it has a mass of \(339.076 \mathrm{~g}\). What is the mole percent of Ne in the gas mixture?

At an underwater depth of \(250 \mathrm{ft}\), the pressure is 8.38 atm. What should the mole percent of oxygen be in the diving gas for the partial pressure of oxygen in the mixture to be 0.21 atm, the same as in air at 1 atm?

A quantity of \(\mathrm{N}_{2}\) gas originally held at 5.25 atm pressure in a 1.00-L container at \(26^{\circ} \mathrm{C}\) is transferred to a 12.5-L container at \(20^{\circ} \mathrm{C}\). A quantity of \(\mathrm{O}_{2}\) gas originally at \(5.25 \mathrm{~atm}\) and \(26^{\circ} \mathrm{C}\) in a 5.00-L container is transferred to this same container. What is the total pressure in the new container?

The planet Jupiter has a surface temperature of \(140 \mathrm{~K}\) and a mass 318 times that of Earth. Mercury (the planet) has a surface temperature between \(600 \mathrm{~K}\) and \(700 \mathrm{~K}\) and a mass 0.05 times that of Earth. On which planet is the atmosphere more likely to obey the ideal-gas law? Explain.

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