A 6.0-cm-diameter cylinder of nitrogen gas has a 4.0-cm-thick movable copper pis
ID: 1331584 • Letter: A
Question
A 6.0-cm-diameter cylinder of nitrogen gas has a 4.0-cm-thick movable copper piston. The cylinder is oriented vertically, as shown in the figure, and the air above the piston is evacuated. When the gas temperature is 15 ?C , the piston floats 20cm above the bottom of the cylinder.
(Figure 1)
A) What is the gas pressure?
B) How many gas molecules are in the cylinder?
C) Then 2.5 J of heat energy are transferred to the gas. What is the new equilibrium temperature of the gas in ?C?
D) What is the final height of the piston?
E) How much work is done on the gas as the piston rises?
Explanation / Answer
a)
The pressure is,
P = mg / A
= density* volume * g
= density * h * A * g / A
= density *height * g
= 8960 * 0.04 * 9.8
= 3512 Pascal
(b)
The ideal gas equation is used to calculate the number of molecules.
p V = n R T
n = p V / RT
=p*pi* r2 H
=3512* pi * 0.03^2* 0.20 /8.31 * 288
= 8.29x10-4 moles
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(c)
For a diatomic gas, with heat added at constant pressure
Heat added = (7/2) n R delta T
2.0 = 3.5 *8.29x10-4 moles * 8.31 * delta T
Delta T = 103.68 C
The new equilibrium temp = 15+103.68 C
= 118.68 C
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d)
The area is
A = pi r^2 = 0.002827 m2
Work done = p delta V and work done = n R delta T so
p delta V =n R delta T
p A delta h = n R delta T
3512 * 0.002827 * delta h = 8.29x10-4 * 8.31 * 103.68 C
deltah = 0.072 meters = 7.2 cm
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e) Work done on piston = -m g delta h
= -volume * density * g * delta h
=- 0.04 * pi * 0.03^2 * 8960 * 9.8 *0.072
= -0.715 J
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