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Jan. 31, 2023•0 likes•2 views

Jan. 31, 2023•0 likes•2 views

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2. A compressed air tank carried by a scuba diver has a volume of 8.0L and a pressure of 140 atm at 20.0Â°C. What is the volume in the tank (in liters) at STP? (5) 3. If 20.0 g of N2 () has a volume of 40 L and a pressure of 4560 mmHg, what is the Celsius temperature? (5) 4. An unknown gas is placed in a 1.500 L bulb at a pressure of 0.468 atm and a temperature of 22.5Â°C, and is found to weugh0.9847 g. What is the molecular weight of the gas? (5) Solution 2) At STP, T2 = 273.15 K and P2 = 1 atm For fixed amount of gas, P1 * V1 / T1 = P2 * V2 / T2 140 * 8.0 / 293 = 1 * V2 / 273.15 V2 = 1044 L so volume in the tank at STP is 1044 L 3) 20 gm N2 = mass / molar mass = 20 / 28 = 0.714 mole. 1 atm = 760 mm Hg. Ideal gas equation is PV = nRT (4560 / 760) atm * 40 L = 0.714 mole * 0.0821 L-atm / mole K * T T = 4094 K = 4094 - 273 = 3821 oC so celcius temperature is 3821 oC. 4) Let molecular weight of gas is M gm / mole. moles of gas = 0.9847 / M mole. ideal gas equation is PV = nRT 0.468 * 1.500 = (0.9847 / M) * 0.0821 * 295.5 M = 34.03 gm / mole. therefore, molecular weight of gas is 34.03 gm / mole. .

- 1. 2. A compressed air tank carried by a scuba diver has a volume of 8.0L and a pressure of 140 atm at 20.0Â°C. What is the volume in the tank (in liters) at STP? (5) 3. If 20.0 g of N2 () has a volume of 40 L and a pressure of 4560 mmHg, what is the Celsius temperature? (5) 4. An unknown gas is placed in a 1.500 L bulb at a pressure of 0.468 atm and a temperature of 22.5Â°C, and is found to weugh0.9847 g. What is the molecular weight of the gas? (5) Solution 2) At STP, T2 = 273.15 K and P2 = 1 atm For fixed amount of gas, P1 * V1 / T1 = P2 * V2 / T2 140 * 8.0 / 293 = 1 * V2 / 273.15 V2 = 1044 L so volume in the tank at STP is 1044 L 3) 20 gm N2 = mass / molar mass = 20 / 28 = 0.714 mole. 1 atm = 760 mm Hg. Ideal gas equation is PV = nRT (4560 / 760) atm * 40 L = 0.714 mole * 0.0821 L-atm / mole K * T T = 4094 K = 4094 - 273 = 3821 oC so celcius temperature is 3821 oC. 4)
- 2. Let molecular weight of gas is M gm / mole. moles of gas = 0.9847 / M mole. ideal gas equation is PV = nRT 0.468 * 1.500 = (0.9847 / M) * 0.0821 * 295.5 M = 34.03 gm / mole. therefore, molecular weight of gas is 34.03 gm / mole.