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\(4 \mathrm{~g}\) of an ideal gas occupies \(5.6035\) litres of volume at \(546 \mathrm{~K}\) and \(2 \mathrm{~atm}\) pressure. What is its molecular weight? (a) 4 (b) 16 (c) 32 (d) 64

Short Answer

Expert verified
The molecular weight of the gas is 16 g/mol (option b).

Step by step solution

01

Define the Ideal Gas Law

The Ideal Gas Law is given by the equation \( PV = nRT \), where \( P \) is the pressure of the gas in atm, \( V \) is the volume in liters, \( n \) is the number of moles, \( R \) is the ideal gas constant \(0.0821 \text{ atm} \cdot \text{L} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}\), and \( T \) is the temperature in Kelvin.
02

Rearrange for Moles

Rearrange the Ideal Gas Law to solve for the number of moles, \( n \): \( n = \frac{PV}{RT} \).
03

Input Known Values

Substitute the known values into the rearranged formula: \( P = 2 \text{ atm} \), \( V = 5.6035 \text{ L} \), \( R = 0.0821 \text{ atm} \cdot \text{L} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \), and \( T = 546 \text{ K} \).
04

Calculate Moles of Gas

Calculate \( n \) using the values: \[ n = \frac{2 \times 5.6035}{0.0821 \times 546} \approx 0.248 \text{ mol} \]
05

Relate Moles to Molecular Weight

The molecular weight (\( M \)) of the gas can be found by rearranging the formula \( n = \frac{m}{M} \), where \( m \) is the mass of the gas in grams. Solving for \( M \) gives \( M = \frac{m}{n} \).
06

Solve for Molecular Weight

Use the mass \( m = 4 \text{ g} \) and \( n \approx 0.248 \text{ mol} \) to find \( M \):\[ M = \frac{4}{0.248} \approx 16.13 \text{ g/mol} \]
07

Select the Closest Answer

Compare \( M \approx 16.13 \text{ g/mol} \) with the given options: (a) 4, (b) 16, (c) 32, (d) 64. The closest match is 16.

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Key Concepts

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

Molecular Weight Calculation
To understand molecular weight, think of it as the sum of the weights of all the atoms in a molecule. This is crucial for determining the identity of the gas in various conditions. In the exercise, we need to calculate the molecular weight of an ideal gas. We begin with the formula to find moles, which is the usual practice when using the Ideal Gas Law:
  • Re-arrange the Ideal Gas Law: \( n = \frac{PV}{RT} \)
By substituting the values provided (pressure \( P = 2 \text{ atm} \), volume \( V = 5.6035 \text{ L} \), temperature \( T = 546 \text{ K} \)), we find the number of moles to be approximately \( 0.248 \text{ mol} \).
After finding the moles, relate it to molecular weight using:
\[ M = \frac{m}{n} \]where \( m \) is the mass of the gas (\( 4 \text{ g} \)). This step involves dividing the mass by the number of moles to find the molecular weight, which for this exercise, is approximately \( 16.13 \text{ g/mol} \).
Choosing the closest option provided, the correct answer is 16.
Moles and Gas Laws
The amount of substance, in chemistry, is often expressed in terms of moles. A mole is a convenient unit that contains Avogadro's number of entities, which is \( 6.022 \times 10^{23} \) molecules. This aids in simplifying reactions and calculations.
Using the Ideal Gas Law, we connect physical quantities such as pressure, volume, and temperature to derive the number of moles. Specifically, with:
  • \( PV = nRT \)
  • \( n = \frac{PV}{RT} \)
We substitute known values into the equation to calculate moles \( n \), binding microscopic gas behavior to measurable properties like pressure and volume.
This principle not only applies to this problem but provides a foundation for understanding real-world gas behavior, illustrating how a measure of something unseen, like moles, affects observable characteristics.
Gas Constant
The gas constant, denoted as \( R \), is a vital part of the Ideal Gas Law. It bridges the gap between pressure, volume, temperature, and moles. The value of \( R \) can vary based on the units used, but here it is \( 0.0821 \text{ atm} \cdot \text{L} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \).
\( R \) is crucial because it provides a universal factor that connects experimental conditions to theoretical calculations. It ensures that pressure and volume changes relate systematically to temperature and moles.
The constant remains a cornerstone in not only understanding ideal gases but forming the base for deviations observed in non-ideal gases under real conditions.

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Most popular questions from this chapter

A \(200 \mathrm{~mL}\) flask having oxygen at \(220 \mathrm{~mm}\) and a \(300 \mathrm{~mL}\) flask having nitrogen at \(100 \mathrm{~mm}\) are connected in such a way that \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) may combine in their volumes, if temperature is kept constant. Find the total pressure of the gaseous mixture. (a) \(158 \mathrm{~mm}\) (b) \(138 \mathrm{~mm}\) (c) \(148 \mathrm{~mm}\) (d) \(168 \mathrm{~mm}\)

A gas has the van der Waals constants, \(\mathrm{a}=1.49 \mathrm{~L}^{2}\) \(\mathrm{atm} \mathrm{mol}^{-2}\) and \(\mathrm{b}=0.04 \mathrm{~L} \mathrm{~mol}^{-1} .\) Its Boyle's temperature is nearly: (a) \(50^{\circ} \mathrm{C}\) (b) \(354^{\circ} \mathrm{C}\) (c) \(454 \mathrm{~K}\) (d) \(408 \mathrm{~K}\)

Van der Waal equation for \(\mathrm{CH}_{4}\) at low pressure can be given as: (a) \(\mathrm{PV}=\mathrm{RT}-\mathrm{Pb}\) (b) \(\mathrm{PV}=\mathrm{RT}+\frac{\mathrm{a}}{\mathrm{V}}\) (c) \(\mathrm{PV}=\mathrm{RT}-\frac{\mathrm{a}}{\mathrm{V}}\) (d) \(\mathrm{PV}=\mathrm{RT}+\mathrm{Pb}\)

The density of an ideal gas is \(0.03 \mathrm{~g} \mathrm{~cm}^{-3}\), Its pressure is \(106 \mathrm{~g} \mathrm{~cm}^{-1} \mathrm{sec}^{-2}\). What is its \(\mathrm{rms}\) velocity (in \(\left.\mathrm{cm} \mathrm{sec}^{-1}\right) ?\) (a) \(10^{3}\) (b) \(3 \times 10^{4}\) (c) \(10^{8}\) (d) \(10^{4}\)

The ratio of root mean square velocity to average velocity of a gas molecule at a particular temperature is: (a) \(1: 1.086\) (b) \(2: 1.086\) (c) \(1.086: 1\) (d) \(1.086: 2\)

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