The diagram below show a container of neon gas. Air outside the container is at

Question

The diagram below show a container of neon gas. Air outside the container is at 1.00 atm. The top lid of the container is a massless piston connected to a spring. When the spring is in its resting position, the volume of the container is 3.00 L. The piston is circular with a diameter of 5.00 cm. When the neon in the container is at 30 C, the spring is at its resting position, and when the temperature is increased to 150 C, the spring compresses 1.30 cm. (a) What is the spring constant? (b) The temperature of the neon is raised to 250 C. What is the new compression of the spring?

Explanation / Answer

When spring is in resting posiition, pressure on both sides of piston is equal.

3 L = 0.003 m^3

1 atm = 101325 Pa

Piston area A = pi/4*D^2 = 3.14/4 *5^2 = 19.625 cm^2

Mass in cylinder m = PV/(RT)

= 101325*(3*10^-3) / (412*(30+273))

m = 0.00243 kg

Volume change = 19.625*1.3 = 25.5125 cm^3 = 0.0255 L

New volume = 3 + 0.0255 = 3.0255 L

PV = mRT

P*3.0255*10^-3 = 0.00243*412*(150+273)

P = 140 kPa

Doing force balance on both sides of piston

140*10^3 *19.625*10^-4 = k*(1.3*10^-2) + 101325*19.625*10^-4

Solving, k = 5834.5 N/m

b)

Assuming x = new compresion

Doing force balance

P*19.625*10^-4 = 5834.5*x + 101325*19.625*10^-4

But P = mRT / V

and new V = 0.003 + 19.625*10^-4 *x

0.00243*412*(250+273) *19.625*10^-4 / (0.003 + 19.625*10^-4 *x) = 5834.5*x + 101325*19.625*10^-4

Solving this, 1.0276 / (0.003 + 19.625*10^-4 *x) = 5834.5*x + 198.85

or x = 0.0895 m or 89.5 cm

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