Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

7. Gases

The Ideal Gas Law: Density

General Chemistry

7. Gases

The Ideal Gas Law: Density

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Multiple Choice

At what temperature in Kelvin does sulfur hexafluoride (SF6) have a density of 0.5550 g/L at a pressure of 0.8210 atm?

350 K

320 K

298 K

273 K

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Verified step by step guidance

Start by using the ideal gas law equation, which is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin.

Rearrange the ideal gas law to solve for temperature \( T \): \( T = \frac{PV}{nR} \).

To find \( n \), the number of moles, use the formula \( n = \frac{m}{M} \), where \( m \) is the mass and \( M \) is the molar mass of sulfur hexafluoride (SF6). The molar mass of SF6 is approximately 146.06 g/mol.

Since density \( d \) is given as 0.5550 g/L, use the relationship \( d = \frac{m}{V} \) to express mass \( m \) in terms of density and volume: \( m = dV \). Substitute \( m = dV \) into \( n = \frac{m}{M} \) to get \( n = \frac{dV}{M} \).

Substitute \( n = \frac{dV}{M} \) into the rearranged ideal gas law equation \( T = \frac{PV}{nR} \) to get \( T = \frac{PV}{\frac{dV}{M}R} \). Simplify to \( T = \frac{PM}{dR} \). Now, plug in the values: \( P = 0.8210 \) atm, \( M = 146.06 \) g/mol, \( d = 0.5550 \) g/L, and \( R = 0.0821 \) L atm/mol K, and solve for \( T \).