The Ideal Gas Law quiz #2 Flashcards
The Ideal Gas Law quiz #2
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By what factor does the gas temperature change if the volume and pressure are both doubled?
If both volume and pressure are doubled, temperature increases by a factor of four (T2 = 4T1). What is the pressure of the gas after it is compressed and cooled?
Pressure decreases if temperature decreases and volume decreases, but exact value requires calculation using PV = nRT. Which equation agrees with the ideal gas law?
PV = nRT Which statement can best be concluded from the ideal gas law?
Pressure, volume, temperature, and moles of gas are all interrelated. How many moles of oxygen (O2) are present in 33.6 L of the gas at 1 atm and 0°C?
Moles = 33.6 L / 22.4 L/mol = 1.5 mol. At 600 K, how many moles of gas are in a 1.00 L cylinder at atmospheric pressure?
n = PV/RT = (1 × 1) / (0.08206 × 600) ≈ 0.0203 mol. Which of the following would behave the most like an ideal gas when confined to a 5.0 L container?
Helium (He) would behave most ideally. How many moles of chlorine gas at 110°C and 28.0 atm would occupy a vessel of 15.0 L?
Convert 110°C to K: 383 K. n = PV/RT = (28 × 15) / (0.08206 × 383) ≈ 13.4 mol. What quantity in moles of chlorine gas at 120.0°C and 33.3 atm would occupy a vessel of 12.0 L?
Convert 120°C to K: 393 K. n = (33.3 × 12) / (0.08206 × 393) ≈ 12.3 mol. Assume Patm = 1.00 atm. What is the gas pressure Pgas if the liquid level is equal inside and outside?
Pgas = Patm = 1.00 atm. What is the mass of 8.55 L of phosphine gas (PH3) at STP? Round to 3 significant figures.
Moles = 8.55/22.4 ≈ 0.382 mol. Mass = 0.382 × 34.0 g/mol ≈ 13.0 g. What is the pressure in a 5.80 L container that has 0.545 moles of oxygen gas at 22.0°C?
Convert 22°C to K: 295 K. P = nRT/V = (0.545 × 0.08206 × 295) / 5.80 ≈ 2.27 atm. What is the temperature (in K) of 0.300 mole of neon in a 2.00 L vessel at 4.68 atm?
T = PV/nR = (4.68 × 2.00) / (0.300 × 0.08206) ≈ 380 K. How many molecules of N2 are in a 400.0 mL container at 780 mm Hg and 135°C?
Convert 400 mL to 0.400 L, 780 mm Hg to 1.03 atm, 135°C to 408 K. n = PV/RT = (1.03 × 0.400) / (0.08206 × 408) ≈ 0.0125 mol. Molecules = 0.0125 × 6.022×10^23 ≈ 7.53×10^21. How many moles of CO are contained in a 5.00 L tank at 155°C and 2.80 atm?
Convert 155°C to 428 K. n = (2.80 × 5.00) / (0.08206 × 428) ≈ 0.396 mol. What is the pressure in a 9.00 L tank with 75.1 grams of fluorine gas at 355 K?
Moles = 75.1/38.0 ≈ 1.98 mol. P = nRT/V = (1.98 × 0.08206 × 355) / 9.00 ≈ 6.41 atm. What pressure will be exerted by 1.5 moles of neon gas occupying 5.0 L at 273 K?
P = nRT/V = (1.5 × 0.08206 × 273) / 5.0 ≈ 6.72 atm. How many molecules of O2 occupy a volume of 1.0 L at 65°C and 778 mm Hg?
65°C = 338 K, 778 mm Hg = 1.02 atm. n = (1.02 × 1.0) / (0.08206 × 338) ≈ 0.0364 mol. Molecules = 0.0364 × 6.022×10^23 ≈ 2.19×10^22. What is the volume of 0.200 mol of an ideal gas at 200 kPa and 400 K?
Convert 200 kPa to 1.97 atm. V = nRT/P = (0.200 × 0.08206 × 400) / 1.97 ≈ 3.33 L. What volume would be occupied by two moles of propane gas using the ideal gas law at STP?
At STP, 1 mole = 22.4 L. Two moles = 44.8 L. How many moles of gas are present if P = 11 atm, V = 1 L, T = 273 K?
n = PV/RT = (11 × 1) / (0.08206 × 273) ≈ 0.492 mol. Which statement best describes the gas in a scuba tank?
The gas in a scuba tank is under high pressure, allowing more moles to be stored in a small volume. If a gas has a volume of 750 L at 298 K and 1 atm, how many moles are present?
n = PV/RT = (1 × 750) / (0.08206 × 298) ≈ 30.7 mol. A fixed container containing an ideal gas is heated. The pressure of the gas increases because
Heating increases kinetic energy, causing more frequent and forceful collisions with the container walls. A sample of O2 occupies 75 L at STP. How many moles are present?
Moles = 75 / 22.4 ≈ 3.35 mol. A real gas will behave most like an ideal gas under conditions of ________.
High temperature and low pressure. How many moles of a gas sample are in a 5.0 L container at 373 K and 203 kPa?
203 kPa = 2.00 atm. n = (2.00 × 5.0) / (0.08206 × 373) ≈ 0.326 mol. The amount of gas that occupies 60.82 L at 31.0°C and 367 mm Hg is ________ mol.
31°C = 304 K, 367 mm Hg = 0.483 atm. n = (0.483 × 60.82) / (0.08206 × 304) ≈ 1.20 mol. Calculate the volume occupied by 2.0 moles of an ideal gas at 25°C and 380 mm Hg.
25°C = 298 K, 380 mm Hg = 0.5 atm. V = nRT/P = (2.0 × 0.08206 × 298) / 0.5 ≈ 97.7 L. If 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy ________ L at STP.
Ratio: (129.3/50.75) × 10.0 = 25.5 L. An 80 g sample of a gas was heated from 25°C to 225°C at constant volume. What happens to the pressure?
Pressure increases as temperature increases. Which describes the volume of 1 mole of gas at STP?
1 mole of an ideal gas occupies 22.4 L at STP. Nitrogen will behave most like an ideal gas under what conditions?
High temperature and low pressure. The pressure of a sample of CH4 gas (6.022 g) in a 30.0 L vessel at 402 K is ________ atm.
Moles = 6.022/16.0 = 0.376 mol. P = nRT/V = (0.376 × 0.08206 × 402) / 30.0 ≈ 0.414 atm. Avogadro's law states that at a given temperature and pressure, what quantity is constant?
The ratio of volume to moles (V/n) is constant. If a sealed, rigid container's pressure is doubled, what will happen to temperature?
Temperature will also double. What is the pressure, in atm, in a 4.00 L tank with 6.65 moles of nitrogen at 69.6°C?
69.6°C = 342.8 K. P = nRT/V = (6.65 × 0.08206 × 342.8) / 4.00 ≈ 46.9 atm. A gas contained within a piston-cylinder assembly is compressed at constant temperature. What happens to pressure?
Pressure increases as volume decreases. What is the value and units of the ideal gas law constant R used in calculations?
R = 0.08206 L·atm/(mol·K) A sample of gas expands from 1.0 m^3 to 4.0 m^3 at constant temperature. What happens to pressure?
Pressure decreases to one-fourth its original value.
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