A mixture containing 0.765 mol \(\mathrm{He}(g), 0.330 \mathrm{mol} \mathrm{Ne}(g),\) and 0.110 \(\mathrm{mol} \mathrm{Ar}(g)\) is confined in a \(10.00-\mathrm{L}\) vessel at \(25^{\circ} \mathrm{C}\) . (a) Calculate the partial pressure of each of the gases in the mixture. (b) Calculate the total pressure of the mixture.

Short Answer

Expert verified
The partial pressures are \(P_{He} = 1.878\: atm\), \(P_{Ne} = 0.813\: atm\), and \(P_{Ar} = 0.270\: atm\). The total pressure of the mixture is \(P_{total} = 2.961\: atm\).

Step by step solution

01

Write down the Ideal Gas Law formula and known values

We know the Ideal Gas Law formula is: \[PV = nRT\] Where: P = Pressure (atm) V = Volume (L) n = moles of the gas R = Ideal Gas constant (0.0821 L atm/mol K) T = Temperature (K) Given values: - Volume (V) = 10.00 L - Temperature (T) = 25°C = (25+273)K = 298 K - moles of He = 0.765 mol - moles of Ne = 0.330 mol - moles of Ar = 0.110 mol
02

Find the partial pressures for He, Ne, and Ar

Calculate the partial pressure for each gas using the Ideal Gas Law formula: \(He: P_{He}=\frac{n_{He}RT}{V}=\frac{(0.765\: mol)(0.0821\: L\: atm/mol\: K)(298\: K)}{10.00\: L}\) \(P_{He} = 1.878\: atm\) \(Ne: P_{Ne}=\frac{n_{Ne}RT}{V}=\frac{(0.330\: mol)(0.0821\: L\: atm/mol\: K)(298\: K)}{10.00\: L}\) \(P_{Ne} = 0.813\: atm\) \(Ar: P_{Ar}=\frac{n_{Ar}RT}{V}=\frac{(0.110\: mol)(0.0821\: L\: atm/mol\: K)(298\: K)}{10.00\: L}\) \(P_{Ar} = 0.270\: atm \)
03

Calculate the total pressure of the mixture

Now that we have the partial pressures for He, Ne, and Ar, we can find the total pressure of the mixture by adding these partial pressures: \(P_{total} = P_{He} + P_{Ne} + P_{Ar} = 1.878\: atm + 0.813\: atm + 0.270\: atm\) \(P_{total} = 2.961\: atm\) So, the total pressure of the mixture is 2.961 atm.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas. It is expressed as the formula: \[\begin{equation}PV = nRT\text{,}\text{where}\begin{align*}&P \text{ is the pressure of the gas in atmospheres (atm),}&V \text{ is the volume in liters (L),}&n \text{ is the number of moles of gas,}&R \text{ is the ideal gas constant, and}&T \text{ is the absolute temperature in kelvin (K).}\end{align*}\text{}\end{equation}\]The law assumes the gas being measured behaves 'ideally,' which means there are no forces between the gas particles and the particles themselves do not take up any space. While no gas is perfectly ideal, many gases at high temperature and low pressure behave closely enough to ideal to use the law for practical calculations.In our exercise, the Ideal Gas Law is used to find the partial pressure of three different gases in a mixture. By knowing the amount in moles of each gas, the volume of the container, the gas constant (R), and the temperature, we can calculate the partial pressure contributed by each gas in the mixture.

Molar Volume
Molar volume is the volume one mole of gas occupies at a given temperature and pressure. It is a useful concept when discussing gases at standard temperature and pressure (STP). For an ideal gas at STP, which is 0°C and 1 atmosphere of pressure, the molar volume is approximately 22.4 liters.In the case of gases that don't strictly follow ideal behavior, calculations involving molar volume can become complex, but the Ideal Gas Law allows us to use the same general approach for any condition, by incorporating temperature and pressure into the equation.When applying the Ideal Gas Law, knowing the molar volume makes it possible to predict how a given amount of gas will expand or contract under different temperature and pressure conditions. This concept is part of the foundation for understanding how partial pressures are calculated in a mixture of gases.

If the container's volume, like the one in our exercise, is fixed and known, then the molar volume is essentially the volume of the container divided by the amount of moles, allowing us to calculate the pressure for each type of gas when we assume ideal conditions.

Gas Constant (R)
The gas constant (R) is a physical constant that appears in a number of fundamental equations in the physical sciences, such as the Ideal Gas Law. The value of R depends on the units used for pressure, volume, and temperature, and it provides a relationship between these variables for an ideal gas.In the context of the Ideal Gas Law:\[\begin{equation}PV = nRT\text{,}\end{equation}\]R is specifically the ideal gas constant with a value of 0.0821 L atm mol-1 K-1 when using atmospheres for pressure, liters for volume, and kelvin for temperature. This value allows us to seamlessly use the Ideal Gas Law to link the measurable properties of a gas to the amount of substance in moles.

For the gases in our exercise, helium (He), neon (Ne), and argon (Ar), the gas constant provides a bridge to finding their respective partial pressures within the fixed volume and at the observed temperature.

Temperature Conversion
Temperature conversion in gas law calculations is important because the Ideal Gas Law requires the use of absolute temperature, which is measured in kelvin (K). The Celsius (°C) scale, commonly used in everyday life and often in laboratory settings, is not an absolute scale and therefore must be converted to kelvin.The conversion is straightforward: the temperature in kelvin is equal to the temperature in Celsius plus 273.15.\[\begin{equation}T(K) = T(°C) + 273.15\text{.}\end{equation}\]This conversion takes into account that the zero on the kelvin scale represents absolute zero, the theoretical point where particles have minimum thermal motion. In the exercise we looked at, the temperature was given as 25°C. Using the conversion formula, we obtained a temperature of 298 K (25°C + 273.15). It is essential to use kelvin in gas calculations because the volume and pressure of a gas are directly proportional to its temperature, but only when measured on an absolute scale like the kelvin scale.

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Most popular questions from this chapter

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