1. Thermal Physics
1.1 Change of Pressure with Altitude and Temperature
p =
Given, T = T0 –az; (K) for z < 11,000 m
a = 0.00649 K/m
a) From the equations of the aerostatic and the perfect gas law, we can write,
p = ρ R T; where P is the atmospheric pressure, R is the gas constant, T is the absolute temperature.
dp = ρgdz; where, ρ is the density, g is the acceleration due to gravity and z is the altitude.
The change in temperature with altitude as suggested by NASA can be found from the equation below:
T = T0 –az; (K) for z < 11,000 m; a = 0.00649 K/m
We can write, = [
Or, = = =
Let us assume that the altitude changes from z1 to z2, the pressure changes from p0 to p.
Integrating both sides for a limit of pressure from p0 to p and the range of altitude from z1
to z2,
=
=
=
=
If we consider the change in temperature from z1 to z2 be T0 to T, then it becomes,
=
Or, =
Or, = [Derived]
b) The root mean square of a gas, vrms =
Let at the temperature T1, vrms
drops by 10%.
90% Vrms =
This gives, T/T0 = 0.9
T = 0.9T0
We have, T = T0 –az
Or, z = = 15.4 Km
The required height is 15.4 km from the sea level.
1.2 Flying with Balloon
a) From the p-T-z relation derived in the Q1.1, we can write,
=
Here, a = 0.0649 Km-1
Atmospheric temperature, T0 = (273+25) K
= 298 K,
Total mass of the balloon, m = mass of the people + mass of the balloon
= (3×65 + 80)
= 275 kg
The gravitational pull on the balloon = 275 × 9.81 = 2697.75 N
The balloon has to overcome this drag using its buoyancy force.
Atmospheric pressure, p0 = 100 kPa
Solving the p-T-z equation, T = 315 K
b) We have, T = T0 –az
Or, T = 298 – 0.00649×1000 = 291.51 K; This is the temperature of the hot air.
c) The temperature difference in problem a = 298 –315 = –17 K
b = 298 – 291.51 = 6.49 K.
The trend shows that the temperature of the balloon cools as it rises.
2. Matter
2.1 a) The spectral series of the H-atom are given by,
υ = Rc (1/n’² - 1/n²)
Putting, n’ = n–1
We can write, υ = Rc (1/(n-1) ² - 1/n²)
= Rc
= Rc
= Rc
As n →∞, we can write, 2n and (n-1) →n
Putting these values, υ →2nRc/n4 = 2Rc/n³ [shown]
b) For an electron of charge, e and mass, m, moving at a constant tangential velocity, v, in a circular orbit of radius r around the nucleus, the centripetal acceleration force is Fc =
The force of attraction working on the electron is, Fe =
At a balance condition, Fe = Fc
=
From where we can rearrange, ½ mv² = = K; this the kinetic energy of the electron.
The potential energy of the electron is equal to the coulomb potential working on it.
Thus, U =
Total energy of the electron, E = K + U = –
For an angular velocity of ω, v = ωr.
But, r =
Putting this value and manipulating, we get, E =
c) From a, v = 2Rc/n³ , and from b, E =
according to Bohr postulate, E = hv
= h × 2Rc/n³
Which gives, E =
The Rydberg constant may be determined as, R =
d) Bohr’s quantization principle justifies the procedure applied in part c). The principle states that the energy level in each orbit of an electron is quantized.
e) Here, E =
From equation 1, E =
Equating these two equations, =
We can write, rn
= ; this is the required quantized orbit radius.
f) We know, the angular momentum of an electron moving at constant tangential velocity v in a circular orbit of r, L = mvr.
The wavelength associated with the electron is given by De Broglie equation,
λ =
From which, we can write, mv =
Plugging this value of mv in the equation of angular momentum, L = –(1)
For a standing wave, the circumference = the number of whole wavelengths.
We can consider the number of principle quantum number n as the number of wavelength.
Thus, 2πr = nλ
Or, λ = ; plugging the value of λ into the equation 1, we get,
L = = = nħ where, ħ =
