Question

1) A cylinder is filled with 0.10 mol of an ideal gas at standard temperature and pressure, and a 1.4-kg piston seals the gas in the cylinder (see figure) with a frictionless seal. The trapped column of gas is 2.1-m high. The piston and cylinder are surrounded by air, also at standard temperature and pressure. The piston is released from rest and starts to fall. The motion of the piston ceases after the oscillations stop with the piston and the trapped air in thermal equilibrium with the surrounding air.

1.4 kg

(a) Find the height of the gas column.
m

(b) Suppose that the piston is pushed down below its equilibrium position by a small amount and then released. Assuming that the temperature of the gas remains constant, find the frequency of vibration of the piston.
Hz

2) A helium balloon is used to lift a load of 100 N. The weight of the envelope of the balloon is 45.0 N and the volume of the helium when the balloon is fully inflated is 31.8 m³. The temperature of the air is 0°C and the atmospheric pressure is 1.00 atm. The balloon is inflated with a sufficient amount of helium gas that the net upward force on the balloon and its load is 31.2 N. Neglect any effects due to the changes of temperature as the altitude changes.

(a) How many moles of helium gas are contained in the balloon?
mol

(b) At what altitude will the balloon be fully inflated?
km
(c) Does the balloon ever reach the altitude at which it is fully inflated?

Yes/No   

(d) If the answer to Part (c) is "Yes," what is the maximum altitude attained by the balloon? (If the answer to Part (c) is "No", enter no.)
km

Answer 1

One question at a time, please

please let me know if there is any mistake. This question was really tough !!

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(a)

Volume of cylinder = hA

pressure of the gas in the cylinder is given as

P = nRT / V = nRT / hA

we need to consider total pressure ( pressure due to atmosphere)

P = Patm + mg / A

nRT / hA = Patm + mg / A

solve this equation for h, we have

at STP conditions, 0.1 mol means 2.24e-3 m³ volume

so,

hA = 2.24e-3

A = 2.24e-3 / 2.1

A = 1.0666e-3 m²

therefore,

h = 2.1 / ( 1 + (mg/ APatm))

h = 2.1 / ( 1 + ( 1.4 * 9.8 / 1.0666e-3 * 1.013e5))

h = 1.863 m

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(b)

Now, we need to apply Newton's second law to the piston

Note that pressure inside the cylinder pushes the piston upwards and atmospheric pressure pushes the piston downwards. The weight of piston acts downwards.

Assuming downward direction as negative.

so,

PA - mg - Patm * A = 0

for small displacmeent, x we have

P'A - mg - Patm = ma

P'A - PA = ma

P' (V+ Ax) = P V

now , using Pressure equation from part (a)

-(nRT / h² ) x = ma

a = - (nRT / mh² ) x

Now, from simple harmonic motion

we know that

a = - w² x

where x is displacement from equilibrium

where w² = k/m

so,

k/m = nRT / mh²

Now,

we know

f = (1/2$\pi$) sqrt ( k/m)

f = (1/2$\pi$) sqrt ( nRT / mh² )

f = (1/2$\pi$) sqrt ( 0.1 * 8.314 * (27 + 273) / 1.4 * 1.863² )

f = 1.14 Hz