The container below holds helium gas at 273 K. The air outside the container is
ID: 1284348 • Letter: T
Question
The container below holds helium gas at 273 K. The air outside the container is at the same temperature, and a pressure of 1.00 atm. The piston is circular, with a mass of 50.0 kg and diameter of 30.0 cm. The spring connecting the piston to the ceiling is at its relaxed length.
(a) Find the pressure and density of the helium. (Take the molar mass of helium to be 4.0 g/mol.)
(b) ln this configuration, the height of the rylindrical volume occupied by the helium is 40 cm. How many moles of helium are present?
(c) The temperature of the helium is raised to 300 K, causing the spring to compress 3.5 cm. What is the spring constant?
Explanation / Answer
Weight of piston = 50*9.81 = 490.5 N
Cross-section area of piston A = 3.14/4 *30^2 = 706.5 cm^2 = 0.07065 m^2
Pressure due to piston weight = 490.5 / 0.07065 = 6942.7 Pa = 6.9427 kPa
Atmospheric pressure = 1 atm = 101.325 kPa
a)
Pressure in container = 101.325 + 6.9427 = 108.268 kPa
Gas constant for Helium R = 8314 / 4 = 2078.5 J/kg-K
Density = P / (RT)
= 108.268*10^3 / (2078.5*273)
= 0.1908 kg/m^3
b)
Volume V = A*height
= 0.07065*0.4
= 0.02826 m^3
PV = nRT
108.268*10^3 *0.02826 = n*8314*273
No of moles n = 0.0013
c)
New volume V2 = 0.07065*(0.4 + 0.035) = 0.03073 m^3
PV/T = constant
108.268*0.02826 / 273 = P2*0.03073 / 300
P2 = 109.4 kPa
Doing force balance across the piston,
109.4*10^3 = 101.325*10^3 + 6.9427*10^3 + k*0.035
Spring constant k = 32438.8 N/m
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