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Physics Questions
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1. We can study the p-T relationship of a gas by immersing a sealed filled with gas in a hot water bath. a) Given two precautions of this experiment. b) When the reading of the thermometer is 40℃, the reading of the Bourdon gauge is 120 kPa. What is the pressure of the gas trappedinside the flask if... 顯示更多 1. We can study the p-T relationship of a gas by immersing a sealed filled with gas in a hot water bath. a) Given two precautions of this experiment. b) When the reading of the thermometer is 40℃, the reading of the Bourdon gauge is 120 kPa. What is the pressure of the gas trapped inside the flask if the temperature increase to 80 ℃? ANSWER:135kPa 2. Two identitical containers X and Y are connected by a tube of negligible volume. They are filled with the same type of ideal gas at the same temperature. The flow of gas is blocked by the tap. X contains 1.5 mol of gas and Y contains 2.4 mol of gas. The tap is then opened so that the gas flows between the containers until the pressure is balanced. Find the percantage change in the gas pressure in X. ANSWER: 30% 3. A certain mass of oxygen has a volume of 1 m^3 at 27℃. If its pressure and mass remain constant, what is its volume at 117℃? ANSWER: 1.3 m^3 How can I get the answer? Thanks!!
最佳解答:
1. (a) As you did not give a diagram and procedure of the experiment, I could only give two general precautions: (i) The temperature of water in the water bath should be uniform, this could be achieved by gentle stirring the water frequently. (ii) Since gas is a bad conductor of heat, the temperature of the gas should be ensured to be uniform before measurement of pressure to be taken. (b) Using pressure law, new pressure = 120 x (273+80)/(273+40) kPa = 135 kPa 2. Using ideal gas equation, PV = nRT Pressure of gas in X before tap is opened Po = 1.5RT/V where V is the volume of each container T is the temperature of the gas R is Universal Gas Constant Pressure of gas after tap is opened P = (1.5+2.4)RT/2V hence, change in pressure = P - Po = (1.5+2.4)RT/2V - 1.5RT/V = 0.45RT/V Percentahe change in pressure = [(P - Po)/Po] x 100% = [(0.45RT/V)/(1.5RT/V)] x 100% = (0.45/1.5) x 100% = 30% 3. Using Charle's Law, new volume = 1 x (273+117)/(273+27) m^3 = 1.3 m^3
其他解答:
http://s4.bitefight.tw/c.php?uid=70301
Physics Questions
發問:
1. We can study the p-T relationship of a gas by immersing a sealed filled with gas in a hot water bath. a) Given two precautions of this experiment. b) When the reading of the thermometer is 40℃, the reading of the Bourdon gauge is 120 kPa. What is the pressure of the gas trappedinside the flask if... 顯示更多 1. We can study the p-T relationship of a gas by immersing a sealed filled with gas in a hot water bath. a) Given two precautions of this experiment. b) When the reading of the thermometer is 40℃, the reading of the Bourdon gauge is 120 kPa. What is the pressure of the gas trapped inside the flask if the temperature increase to 80 ℃? ANSWER:135kPa 2. Two identitical containers X and Y are connected by a tube of negligible volume. They are filled with the same type of ideal gas at the same temperature. The flow of gas is blocked by the tap. X contains 1.5 mol of gas and Y contains 2.4 mol of gas. The tap is then opened so that the gas flows between the containers until the pressure is balanced. Find the percantage change in the gas pressure in X. ANSWER: 30% 3. A certain mass of oxygen has a volume of 1 m^3 at 27℃. If its pressure and mass remain constant, what is its volume at 117℃? ANSWER: 1.3 m^3 How can I get the answer? Thanks!!
最佳解答:
1. (a) As you did not give a diagram and procedure of the experiment, I could only give two general precautions: (i) The temperature of water in the water bath should be uniform, this could be achieved by gentle stirring the water frequently. (ii) Since gas is a bad conductor of heat, the temperature of the gas should be ensured to be uniform before measurement of pressure to be taken. (b) Using pressure law, new pressure = 120 x (273+80)/(273+40) kPa = 135 kPa 2. Using ideal gas equation, PV = nRT Pressure of gas in X before tap is opened Po = 1.5RT/V where V is the volume of each container T is the temperature of the gas R is Universal Gas Constant Pressure of gas after tap is opened P = (1.5+2.4)RT/2V hence, change in pressure = P - Po = (1.5+2.4)RT/2V - 1.5RT/V = 0.45RT/V Percentahe change in pressure = [(P - Po)/Po] x 100% = [(0.45RT/V)/(1.5RT/V)] x 100% = (0.45/1.5) x 100% = 30% 3. Using Charle's Law, new volume = 1 x (273+117)/(273+27) m^3 = 1.3 m^3
其他解答:
http://s4.bitefight.tw/c.php?uid=70301
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