A tank having a volume of 0.1 m^{3} contains helium gas at 150 atm.
How many balloons can the tank blow up, if each filled balloon is a sphere 0.3 m
in diameter at an absolute pressure of 1.2 atm?

Solution:

- Concepts:

The ideal gas law, (Boyle's law) - Reasoning:

At constant temperature P_{1}V_{1 }= P_{2}V_{2}. (Boyle's law)

We are given P_{1}, V_{1}, and P_{2}and are asked to solve for V_{2}. - Details of the calculation:

P_{1 }= 150 atm = 1.515*10^{7 }Pa. V_{1 }= 0.1m^{3}.

P_{2 }= 1.2 atm = 1.212*10^{5 }Pa.

V_{2 }= P_{1}V_{1}/P_{2 }= 12.5 m^{3}.

Let n = number of balloons and V_{b }= the volume of each blown-up balloon.

V_{b }= (4π/3)r^{3 }= 1.414*10^{-2 }m^{3}.

n = V_{2}/V_{b }= 884. The tank can blow up 884 balloons.

A vertical open glass tube of length h is half-submerged in mercury. The top end of the tube is then closed and the tube is slowly pulled out until the bottom of the tube is barely submerged in the mercury. What is the length of the mercury column remaining in the tube? The atmospheric pressure corresponds to the pressure of a column of mercury of height H. Assume the temperature is constant.

Solution:

- Concepts:

The ideal gas law - Reasoning:

At constant temperature PV = constant (Boyle's law).

When a length h/2 of the tube is above the mercury, the volume of the gas in the tube is Ah/2 and the pressure is atmospheric pressure P_{0}.

When a length h of the tube is above the mercury, the mercury rises to a height h_{1}and the gas occupies a volume A(h - h_{1}). Since PV = constant (Boyle's law), the gas pressure is

P = ½P_{0}h/(h-h_{1}).

We have P_{0 }= ½P_{0}h/(h-h_{1}) + ρ_{mercury}gh_{1}. - Details of the calculation:

Given: ρ_{mercury}gH = P_{0}, so P_{0 }= ½P_{0}h/(h-h_{1}) + P_{0}h_{1}/H.

Therefore

½hH_{ }= h_{1}H + h_{1}h - h_{1}^{2}, h_{1}^{2}- h_{1}(H + h) + ½hH = 0, h_{1}= (H + h)/2 – (H^{2}+ h^{2})^{½}/2,

A movable heavy piston is supported by a spring inside a vertical cylindrical container as shown. When all air is pumped out of the container, the piston is in equilibrium, as shown in diagram 1, with only a tiny gap between the piston and the bottom of the container. When a portion of gas at temperature T is introduced under the piston, the piston rises to a height h as shown in diagram 2.

What would be the height of the piston above the bottom of the container if the gas was heated to temperature 2T? Assume the piston moves without friction and that the spring obeys Hooke's law.

Solution:

- Concepts:

The ideal gas law, Hooke's law - Reasoning:

Let k be the spring constant of the spring. In equilibrium PA = kh.

P = pressure, A = area of piston, h = height of the piston above the bottom - Details of the calculation:

P_{1}/P_{2}= h_{1}/h_{2}. (Hooke's law)

From the ideal gas law we have P_{1}V_{1}/T_{1}= P_{2}V_{2}/T_{2}.

Here T is the absolute temperature.

V = hA, so P_{1}h_{1}/T_{1}= P_{2}h_{2}/T_{2},

With T_{2}= 2T_{1}we have P_{2}= 2P_{1}h_{1}/h_{2}, (h_{1}/h_{2})^{2}= ½.

So if the piton rises to a height h when the gas temperature is T,

then it rises to height (√2)h when the gas temperature is 2T.

Near the surface of Earth where the gravitational acceleration has magnitude
g, derive an expression for the atmospheric pressure as a function of altitude y
assuming that

(a) the temperature T is constant.

(b) the temperature T is not constant, it decreases with altitude y. In
the standard atmosphere T = T_{0 }- ay, with a = 0.0065 K/m.

Assume the atmosphere is an ideal gas of molecules with particle mass m.

Solution:

- Concepts:

The ideal gas law - Reasoning:

Let us consider a column of gas containing ρ_{particle}particles of mass m per unit volume near the Earth's surface.The pressure times the area (i.e. the force) at height y must exceed that at height y + dy by the weight of the intervening gas.

We need P_{y+dy}A - P_{y}A = -mρ_{particle}Vg = -mρ_{particle}gAdy, or dP = -mρ_{particle}gdy.

The ideal gas law lets us express ρ_{particle}in terms of the temperature and the pressure. - Details of the calculation:

(a) T is constant.

From the ideal gas law we know that P = ρ_{particle}kT, with k being the Boltzmann constant. We may therefore write

dP = -P(mg/kT)dy, or dP/P = -(mg/kT)dy.

This integrates to

P = P_{0}e^{-mgy/(kT)}.

The pressure decreases exponentially with altitude.

(b) T = T_{0 }- ay. We now have

dP/P = -mgdy/(k(T_{0 }- ay)).

This integrates to

ln(P/P_{0}) =( mg/ka)ln(1 - ay/T_{0}), or

(P/P_{0})^{(ka/mg) }= (1 - ay/T_{0}).

(a) By considering the forces acting on a thin concentric shell of atmosphere at a distance r from the center of the earth, show that the variation of pressure P with height in terms of the local density of the atmosphere ρ and the local gravitational acceleration g is given by

dP/dr = -ρg.

(b) If the temperature is 27^{o} C at the earth's
surface and the atmosphere is assumed to be an ideal gas composed approximately
of 80% diatomic ^{14}N and 20% diatomic ^{16}O, estimate the
vertical distance over which the pressure falls to 1/e of its value at the
surface (this is called the scale height of the atmosphere). To simplify your
estimate you may assume the temperature to remain constant with height above the
surface (which typically introduces an error of ~20% compared to a more
realistic variation of temperature).

Solution:

- Concepts:

Pressure, the ideal gas law - Reasoning:

Consider a volume of air of height h in equilibrium. It does not rise or fall. The net force on it must be zero.

Therefore P_{bottom }- P_{top }= ρhg. The ideal gas law relates the density to the pressure. - Details of the calculation:

(a) Since R_{Earth}>> h, the height of the atmosphere, we may consider the concentric shells locally flat. Then

F_{net }= P_{bottom}A - P_{top}A^{ }- ρAΔrg = 0.

P_{top}- P_{bottom}= -ρΔrg.

dP/dr = -ρg.

(b) dP/dr = -(0.2 m_{O}+ 0.8 m_{N})ρ_{particle}g = -mρ_{particle}g.

m = 0.2*32u + 0.8*28u = 28.8u = 4.784*10^{-26}kg.

But from the ideal gas law we know that P = ρ_{particle}kT, with k being the Boltzmann constant. We may therefore write

dP/dr = -P(mg/kT), or

dP/P = -(mg/kT)dr.

If the temperature T is constant, then this integrates to

P = P_{0}exp(-mg(r-r_{0})/(kT)).

For P/P_{0}= 1/e we need mg(r-r_{0})/(kT) = 1, r – r_{0}= kT/mg.

kT = 1.381*10^{-23}*300 J. r – r_{0}= 8.84 km.

This is much less than R_{Earth}= 6.37*10^{3}km, so the locally flat assumption is justified.

Consider a circular cylinder of radius R and length L, rotating about
its symmetry axis with angular velocity ω and containing an ideal gas with
particles of mass M. We assume that the system is in thermal equilibrium at
temperature T = 300 K, that the gas is at rest in a reference frame rotating
with the cylinder, and that the particle velocities are small enough that we can
ignore all the Coriolis forces.

(a) If P(0) is the pressure on the axis of
rotation, find the pressure P(r) between r = 0 and r = R.

(b) If R = 1 km, the gas consists of Nitrogen molecules, ω = 0.2/s, and P(R)
= 101 kPa, find the pressure on the axis.

(c) If R = 0.1 m, the gas consists of Nitrogen molecules, ω = 0.2/s, and P(R)
= 101 kPa, find the pressure on the axis.

Solution:

- Concepts:

Fictitious forces in an accelerating frame - Reasoning:

Fictitious forces appear in the rotating frame. Consider the gas in a cylindrical shell of radius r and thickness dr. The centrifugal outward force on the gas must be balanced by a force due to a pressure difference. - Details of the calculation:

(a) (P_{out}– P_{in})2πr = ρ2πr dr ω^{2}r

dP/dr = (N/V)Mω^{2}r

Ideal gas law: P = (N/V)kT

Therefore dP/dr = (P/(kT))Mω^{2}r, dP/P = Mω^{2}rdr/(kT).

Integrating we obtain P(r) = P(0)exp(Mω^{2}r^{2}/(2kT)).

(b) M/(2kT) = 28*1.66*10^{-27}/(2*1.38*10^{-23}*300) = 5.61*10^{-6}.

ω^{2}r^{2}= 4*10^{4}**.**exp(Mω

^{2}r^{2}/(2kT)) = 1.25

P(0) = 101kPa/1.25 = 81 kPa.

(c) ω^{2}r^{2}= 4*10^{-4}**.**exp(Mω

^{2}r^{2}/(2kT)) = 2.2*10^{-9 }P(0) ≈ P(R).