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Chapter 10: Problem 45

Which gas is most dense at \(1.00 \mathrm{~atm}\) and \(298 \mathrm{~K} ? \mathrm{CO}_{2}\), \(\mathrm{N}_{2} \mathrm{O}\), or \(\mathrm{Cl}_{2}\). Explain.

Short Answer

Expert verified
The most dense gas at 1.00 atm and 298 K among CO鈧�, N鈧侽, and Cl鈧� is Cl鈧� with a density of 2.915 g/L. This is determined by calculating the molar masses of each gas and using the Ideal Gas Law to find the densities. The order of densities is Cl鈧� (2.915 g/L) > N鈧侽 (1.818 g/L) > CO鈧� (1.817 g/L).

Step by step solution

01

Understand the relationship between pressure, temperature, volume, and density of a gas

Density is defined as mass divided by volume. To find the volume occupied by a given mass of a gas under specific pressure and temperature conditions, we need to use the Ideal Gas Law: \( PV = nRT \), where P is pressure, V is volume, n is the amount of gas in moles, R is the ideal gas constant, and T is temperature. To find the density, we can use the equation \( 蟻 = \frac{m}{V} \), where 蟻 is density, m is mass, and V is volume. Combining the Ideal Gas Law and density equation, we get the following equation: \( 蟻 = \frac{m}{\frac{nRT}{P}} = \frac{mP}{nRT} \)
02

Convert mass to moles and substitute

Since the relationship is more convenient in moles, we can replace mass (m) with the number of moles (n) times the molar mass (M). Our equation now becomes: \( 蟻 = \frac{nMP}{nRT} \) However, 'n' in the numerator and denominator will cancel each other out, leaving us with the following simplified equation for density: \( 蟻 = \frac{MP}{RT} \)
03

Calculate the molar masses

In order to find which gas among CO鈧�, N鈧侽, and Cl鈧� is most dense at 1.00 atm and 298 K, we need to know their respective molar masses. Molar mass of CO鈧� = 12.01 (C) + 2 脳 16.00 (O) = 44.01 g/mol Molar mass of N鈧侽 = 2 脳 14.01 (N) + 16.00 (O) = 44.02 g/mol Molar mass of Cl鈧� = 2 脳 35.45 (Cl) = 70.90 g/mol
04

Calculate the densities at 1.00 atm and 298 K

Given the values P = 1.00 atm, T = 298 K, and R 鈮� 0.0821 atm L/mol K, we can now calculate the density of each gas using the formula \( 蟻 = \frac{MP}{RT} \). Density of CO鈧� = \( \frac{44.01 \text{g/mol} \times 1.00 \text{atm}}{0.0821 \text{atm L/mol K} \times 298 \text{K}} \) = 1.817 g/L Density of N鈧侽 = \( \frac{44.02 \text{g/mol} \times 1.00 \text{atm}}{0.0821 \text{atm L/mol K} \times 298 \text{K}} \) = 1.818 g/L Density of Cl鈧� = \( \frac{70.90 \text{g/mol} \times 1.00 \text{atm}}{0.0821 \text{atm L/mol K} \times 298 \text{K}} \) = 2.915 g/L
05

Compare the densities and draw a conclusion

Comparing the densities of the three gases at 1.00 atm and 298 K, we observe the following order: Cl鈧� (2.915 g/L) > N鈧侽 (1.818 g/L) > CO鈧� (1.817 g/L) Thus, Cl鈧� is the most dense gas at 1.00 atm and 298 K.

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Most popular questions from this chapter

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