A cylinder contains oxygen gas (o2) at a pressure of 2.60 atm. The volume is 6.0

Question

A cylinder contains oxygen gas (o2) at a pressure of 2.60 atm. The volume is 6.00 L, and the temperature is 300 K. Assume that the oxygen may be treated as an ideal gas. The oxygen is carried through the following processes: 1.) Heated at constant pressure from the initial state (state 1) to state 2, which has 420 k. 2.) Cooled at constant volume to 250 k (state 3) 3.) Compressed at constant temperature to a volume of 6.00 L (state 4) 4.) Heated at constant volume to 300 K, which takes the system back to state 1. Calculate Q for each of the four processes. Will rate!!!

Explanation / Answer

A cylinder contains oxygen gas at a pressure of 2.60 atm. The volume is 6.00 L, and the temperature is 300 K. Assume that the oxygen may be treated as an ideal gas. The oxygen is carried through the following processes: Ideal gas equation: Pressure * Volume = (number of moles) * Ideal gas constant * Temperature 2.60 * 6.00 = (number of moles) * 0.08205 * 300 (number of moles) = 0.63376 Volume of cylinder = Area of base * height Pressure = Force ÷ Area of base Pressure * Volume = Force * height Force * change of height = Work So, Pressure * Change of Volume = Work OR Change of Pressure * Volume = Work SO Work = (number of moles) * Ideal gas constant * Change of Temperature Work = change of Energy Energy = (number of moles) * Ideal gas constant * Change of Temperature OR Energy = Change of Pressure * Volume When the temperature changes, the following equation will determine the increase or decrease of heat energy. ?Q = (number of moles) * Ideal gas constant * Change of Temperature ?Q =0.63376 * 0.08205 * ?T When the pressure or volume changes, the following equation will determine the work done. Work = Constant P * ?V OR Work = ?P * Constant V (1) Heated at constant pressure from the initial state (state 1) to state 2, which has 420 K. Since #3 says the volume is compressed back to 6.00 L, the volume must increase as the temperature increases from 300°K to 420°K. New Volume = 6.00 * 420/300 = 8.4 L ?V = 8.4 – 6.00 = 2.4 Work = P * ?V = 2.60 * 2.4 = 6.24 ?Q = 0.63376 * 0.08205 * (420 – 300) = 6.24 (2) Cooled at constant volume to 250 K (state 3). The temperature decreases from 420 to 250, but the volume remains 8.4 L. So the pressure must decrease. New pressure = 2.60 * 250/420 = 1.548 atm Work = ?P * V = (1.548 – 2.60) * 8.4 = -8.84 ?Q =0.63376 * 0.08205 * (250 – 420) = -8.84 (3) Compressed at constant temperature to a volume of 6.00 L (state 4). The pressure must increase as the volume decreases from 8.4 L to 6.00 L. New Pressure * 6 = 1.548 * 8.4 New Pressure = 2.167 ?P = 2.167 – 1.548 = 0.6192 Work = ?P * V = 0.6192 * 8.4 = 5.20 Since the temperature did not change, ?Q = 0

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