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!!
Physics Questions
2010-02-13 4:37 am
回答 (2)
2010-02-13 5:11 am
✔ 最佳答案
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
2010-02-13 4:47 am
收錄日期: 2021-04-29 17:31:39
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