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

Short Answer

Expert verified
Therefore, the density of hydrogen bromide at these conditions is \(3.01 g/L\).

Step by step solution

01

Recalling the Ideal Gas Law

Recall the formula for the ideal gas law, which is \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles of the gas, R is the universal gas constant, and T is the temperature. In this case, we need to find the density (d), which is the mass(m) of the gas divided by the volume (V). So, we should rearrange the formula to find P in terms of density.
02

Change the Units of Given Quantities

First, convert the given pressure from millimeters of mercury (mmHg) to atmosphere (atm), since R's value is usually in atm. 1 atm is equivalent to 760mmHg. Therefore, the pressure in atm will be \(733mmHg * (1atm / 760mmHg) = 0.964 atm\). Next, convert the given temperature from Celsius to Kelvin by adding 273.15 to it. The temperature in Kelvin will then be \(46C + 273.15K = 319.15K\).
03

Rearrange Ideal Gas Law Formula to Find Density

Rearrange the ideal gas law to find formula for density, which is \(d = \frac {m} {V} = \frac {nRT}{P}\). Here, mass m is calculated by multiplying the moles of gas (n) by its molar mass. Hence, the formula becomes \(d = \frac {PM}{RT} = \frac {0.964 atm \times 80.91 g/mol}{0.0821 L atm/mol K \times 319.15 K}\) where M is molar mass of HBr which is 80.91 g/mol and R is the ideal gas constant 0.0821 L atm/mol K.
04

Calculate the Density

Substitute all the known values into the formula and calculate for the density, \(d = \frac {0.964 atm \times 80.91 g/mol}{0.0821 L atm/mol K \times 319.15 K} = 3.01 g/L\).

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Key Concepts

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

Density Calculation
When we talk about the density of a substance, we are referring to how much mass is contained in a specific volume of that substance. For gases, this principle is no different, but calculating the density can often involve understanding and applying the Ideal Gas Law, which is represented by the equation
\(P V = n R T\).
In practical terms, density (\(d\)) of a gas can be found by rearranging the Ideal Gas Law to the form \( d = \frac{m}{V} = \frac{P M}{R T} \), where \(m\) is mass, \(V\) is volume, \(P\) is pressure, \(M\) is the molar mass of the gas, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin. By rearranging the equation in this way, we derive a direct relationship between the density of a gas and the conditions such as pressure and temperature in which it is contained.
The Gas Constant (R)
When dealing with gas calculations, one of the most fundamental constants you'll encounter is the ideal gas constant, denoted as \(R\). The value of \(R\) can be expressed in various units, but the most common one used for calculations involving atm, L, mol, and K is \(0.0821 L\cdot atm/(mol\cdot K)\). This constant is a part of the Ideal Gas Law and plays a crucial role in relating pressure, volume, temperature, and the number of moles of a gas. It's important to remember that \(R\)'s value needs to be consistent with the units used for pressure, volume, and temperature to ensure correct calculations.
Converting Units
In the realm of chemistry and physics, being adept at converting units is an essential skill. For gases, this becomes particularly important because the Ideal Gas Law involves units for pressure (P), volume (V), and temperature (T) that must be compatible.
To accurately calculate the density of a gas, as seen in our example, the pressure in mmHg must be converted to atm, and the temperature from Celsius to Kelvin. This is because the value of the ideal gas constant (R) is usually given in units of atm, L, mol, and K, therefore, having consistent units throughout the calculation is vital.
As a quick reference:
  • To convert pressure from mmHg to atm, divide by 760 (since 760 mmHg = 1 atm).
  • To convert temperature from Celsius to Kelvin, add 273.15.
This ensures that all the values are compatible when plugged into the Ideal Gas Law or any related formula.
Molar Mass
Molar mass, often represented with the symbol \(M\), is a key factor when calculating the density of a gas using the Ideal Gas Law. It is defined as the mass of one mole of a substance and is typically expressed in grams per mole (g/mol).
Knowing the molar mass of a compound like hydrogen bromide (HBr), which is 80.91 g/mol, allows us to relate the mass of the gas to the number of moles present. This information is crucial when rearranging the Ideal Gas Law to solve for density. The molar mass becomes an intermediary variable that bridges moles and mass, making it possible to calculate density if the other conditions (pressure, volume, and temperature) are known.
In practice, the molar mass is obtained from the periodic table by adding the atomic masses of the constituent elements in a compound (1 mole of H + 1 mole of Br for HBr).

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Most popular questions from this chapter

In 2012 , Felix Baumgartner jumped from a balloon roughly \(24 \mathrm{mi}\) above Earth, breaking the record for the highest skydive. He reached speeds of more than 700 miles per hour and became the first skydiver to exceed the speed of sound during free fall. The helium-filled plastic balloon used to carry Baumgartner to the edge of space was designed to expand to \(8.5 \times 10^{8} \mathrm{~L}\) in order to accommodate the low pressures at the altitude required to break the record. (a) Calculate the mass of helium in the balloon from the conditions at the time of the jump \((8.5 \times\) \(\left.10^{8} \mathrm{~L},-67.8^{\circ} \mathrm{C}, 0.027 \mathrm{mmHg}\right) .\) (b) Determine the volume of the helium in the balloon just before it was released, assuming a pressure of 1.0 atm and a temperature of \(23^{\circ} \mathrm{C}\).

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