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At \(27^{\circ} \mathrm{C}, 10.0\) moles of a gas in a \(1.50-\mathrm{L}\) container exert a pressure of 130 atm. Is this an ideal gas?

Short Answer

Expert verified
No, the gas is not ideal because the calculated \( PV \) differs from \( nRT \).

Step by step solution

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01

Understand the Ideal Gas Law

The ideal gas law is written as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
02

Convert Temperature to Kelvin

The temperature is given in Celsius. Convert it to Kelvin by adding 273.15: \[ T = 27^{\circ}C + 273.15 = 300.15 \text{ K} \]
03

Insert Values into the Ideal Gas Law

In the ideal gas law formula \( PV = nRT \), insert the given values: \[ P = 130 \text{ atm}, \quad V = 1.50 \text{ L}, \quad n = 10.0 \text{ moles}, \quad T = 300.15 \text{ K} \]Use \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \).
04

Calculate the Right Side of the Equation

Calculate \( nRT \):\[ nRT = 10.0 \text{ mol} \times 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \times 300.15 \text{ K} = 246.67665 \text{ L atm} \]
05

Calculate the Left Side of the Equation

Calculate \( PV \):\[ PV = 130 \text{ atm} \times 1.50 \text{ L} = 195 \text{ L atm} \]
06

Compare PV and nRT

Compare the calculated values of \( PV \) and \( nRT \):* \( PV = 195 \text{ L atm} \)* \( nRT = 246.67665 \text{ L atm} \)Since \( PV eq nRT \), the gas does not behave as an ideal gas under these conditions.

Key Concepts

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

Pressure
Pressure is a force applied over a certain area. In terms of gases, it's the force that the gas molecules exert when they collide with the walls of their container.
When dealing with the ideal gas law, the pressure must be in a standardized unit, such as atmospheres (atm). This is because it needs to correspond with the other constants used in calculations, like the gas constant.
In our exercise, we see how a gas in a container with a specific volume (1.50 L) exerts a particular amount of pressure (130 atm). The force exerted by the gas particles is affected by various factors including temperature and the number of molecules present.
  • Higher temperature can increase pressure because particles move faster and collide more.
  • Adding more moles of gas increases pressure due to more particle collision with walls.
  • Pressure and volume are inversely related if temperature and moles of gas are constant (Boyle’s Law).
Temperature conversion
Often, temperature needs to be converted from Celsius to Kelvin in chemistry problems that involve gas laws. Kelvin is the standard temperature scale used because it starts at absolute zero, the theoretically coldest point where all molecular motion stops, ensuring all values are positive.
The conversion is simple: you add 273.15 to the Celsius temperature. This ensures the accuracy needed in calculations involving gas laws like the Ideal Gas Law.
For our example, the temperature of 27°C was converted as follows:
  • Formula: \[ T(K) = T(°C) + 273.15 \]
  • Converted: \[ 27°C + 273.15 = 300.15 ext{ K} \]
Gas constant
The gas constant, denoted as \( R \), is a crucial component of the ideal gas law equation \( PV = nRT \). It serves as a bridge, allowing different units to fit together in a single equation.
For ideal gas calculations, \( R \) typically equals 0.0821 L atm K^{-1} mol^{-1}. This value ensures consistency among variables (pressure, volume, temperature in Kelvin, moles) in calculations.
Adjusting \( R \) would allow you to use different units for pressure or volume if needed, but it's essential to maintain consistency for accurate calculations. This is why it's vital to check unit compatibility when solving problems using the ideal gas law.
Moles
Moles are a basic unit in chemistry representing a quantity of substance. Specifically, one mole contains Avogadro's number of particles, which are approximately \( 6.022 imes 10^{23} \) particles of the substance.
In gas law calculations, moles indicate the amount of gas present. This is crucial because it directly influences the pressure exerted by the gas in a container. Knowing the number of moles helps predict how the gas will behave under different conditions based on the Ideal Gas Law.
In our scenario, we are given 10.0 moles of gas. Alongside pressure (130 atm), volume (1.50 L), and temperature (300.15 K), these moles help us determine whether the gas behaves ideally or not, by comparing actual pressure-volume work against theoretical calculations.

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Most popular questions from this chapter

The apparatus shown here can be used to measure atomic and molecular speeds. Suppose that a beam of metal atoms is directed at a rotating cylinder in a vacuum. A small opening in the cylinder allows the atoms to strike a target area. Because the cylinder is rotating, atoms traveling at different speeds will strike the target at different positions. In time, a layer of the metal will deposit on the target area, and the variation in its thickness is found to correspond to Maxwell's speed distribution. In one experiment it is found that at \(850^{\circ} \mathrm{C}\) some bismuth (Bi) atoms struck the target at a point \(2.80 \mathrm{~cm}\) from the spot directly opposite the slit. The diameter of the cylinder is \(15.0 \mathrm{~cm},\) and it is rotating at 130 revolutions per second. (a) Calculate the speed (in \(\mathrm{m} / \mathrm{s}\) ) at which the target is moving. (Hint: The circumference of a circle is given by \(2 \pi r\), where \(r\) is the radius.) (b) Calculate the time (in seconds) it takes for the target to travel \(2.80 \mathrm{~cm} .\) (c) Determine the speed of the \(\mathrm{Bi}\) atoms. Compare your result in part (c) with the \(u_{\mathrm{rms}}\) of \(\mathrm{Bi}\) at \(850^{\circ} \mathrm{C}\). Comment on the difference.

The empirical formula of a compound is \(\mathrm{CH}\). At \(200^{\circ} \mathrm{C}\) \(0.145 \mathrm{~g}\) of this compound occupies \(97.2 \mathrm{~mL}\) at a pressure of \(0.74 \mathrm{~atm}\). What is the molecular formula of the compound?

Write the van der Waals equation for a real gas. Explain the corrective terms for pressure and volume.

The following procedure is a simple though somewhat crude way to measure the molar mass of a gas. A liquid of mass \(0.0184 \mathrm{~g}\) is introduced into a syringe like the one shown here by injection through the rubber tip using a hypodermic needle. The syringe is then transferred to a temperature bath heated to \(45^{\circ} \mathrm{C},\) and the liquid vaporizes. The final volume of the vapor (measured by the outward movement of the plunger) is \(5.58 \mathrm{~mL},\) and the atmospheric pressure is \(760 \mathrm{mmHg}\). Given that the compound's empirical formula is \(\mathrm{CH}_{2}\), determine the molar mass of the compound.

Calculate the density of hydrogen bromide (HBr) gas in \(\mathrm{g} / \mathrm{L}\) at \(733 \mathrm{mmHg}\) and \(46^{\circ} \mathrm{C}\).

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