KTG Flashcards
Boyle’s Law
Volume of a given mass of a gas is inversely proportional to its pressure (PV = K)
Charles’ Law
Between volume and temperature at constant pressure (V proportional T)
Gay Lussac’s Law
Between Pressure and temperature (P proportional to T)
Derive Perfect gas equation
PV = nRT
PV = kNT
P1V1/T1 = P2V2/T2
Value of R
8.31J / mole
Value of k
1.38 * 10^-23
Explain R - P graph
- For ideal gas R = 8.31 and is constant
- For real gases, higher temperature = lower R value
- Higher P = Higher R value
Explain P V graph
- For ideal gas PV curve more to the right (Predicted by Boyle’s Law)
Explain TV graph
- Higher P has higher value of graph
- Explained by Charles’ Law
Postulates of KTG
- All gases consist of molecules
- Size of molecules negligible compared with distance between
- Molecules in state of continuuous random motion
- Molecules collide with one another
- Collisions are perfectly elastic (no forms of attraction)
- Between two collisions a molecule moves in a straight path
- Collisions are almost instanteous
- Density remains uniform throughout
Expression for pressure exerted by gas
P = 1/3 * ro * v^2
Note: v here is mean v or rms v
Derive it
Relation between pressure and KE per unit volume
P = 2/3 * avg KE per unit volume
Relation between P and KE
PV = 2/3 * avg KE
Vrms in terms of density
v = root (3P / ro)
Formula of energy for gas
E = 3/2 RT {For one mole}
E = 3/2 kT {Per molecule}
v rms formula (in terms of R)
v rm = root (3RT / M)
How to calculate pressure in a mercury tube
P = h ro g
Derivation of Boyle’s Law, Charles’ Law, Gay Lussac’s Law, etc.
Derive it
Average Speed
Arithmetic mean of speed of molecules of a gas at a given temperature
Root Mean square speed
Defined as the square root of the mean of the squares of the speed of the molucles of a gas
Most probable speed
Speed possessed by the maximum number of molecules in a gas sample at a given temperature
Relation between v(rms), v(mean, v(mp)
Vrms > V mean > V mp
Degrees of freedom
Total number of co-ordinaes or independent quantities required to describe completely the position and configuration of the system
Degrees of freedom of a system formula
3N - k
N is number of particles in system
k is number of independent relations betweeen particles
Length of Hg Tube
13.6m
Value of pressure in Pa at STP
P = 1atm = 1.013 * 10^5
Mixture of hydrogen and oxygen has volume 2000 cm^3, temp 300K, pressure 100 kPa and mass 0.76g. Find ratio of no. of moles of hydrogen to no. of moles of oxygen in mixture
PV = nRT
n = PV / RT = 0.08
n1 +n2 = 0.08
n1(2) +n2(32) = 0.76
Hence, n1/n2 = 3/1
Degrees of Freedom of:
1. Rigid Body
2. Monoatomic Gas
3. Diatoomic gas
4. Triatomic Gas (non - linear)
5. Triatomic (linear)
- 6 (3 translatory and 3 rotational)
- 3 (Only translatory)
- 5
- 6
- 7
Law of Equipartition of Energy
In any dynamical system in thermal equilibrium, the energy is equally distributed among its various degrees of freedom and energy associated with each degree of freedom per molecule is 1/2 kT
Law of Equipartition of Energy derivation
1/2 mv^2 = 3/2 kT
If v^2 is the mean squared of components vx^2 +vy^2 + vz^2
1/2 mvx^2 + 1/2 mvy^2 + 1/2 mvz^2 = 3/2 kbT
Hence, since 1/2 mvx^2 = 1/2 mvy^2 = 1/2 mvz^2,
Avg. KE per molecule per degree of freedom is 1/2 kT
Specific Heats of:
1. Monoatomic
2. Diatomic
3. Triatomic gases
- Cv = 3/2 RT (Cp = 5/2RT)
- Cv = 7/2 RT (Cp = 7/2 RT)
- Cv = 3R (Cp = 4R)
4.
Relation between gamma (Cp / Cv) and f
gamma = 1 + 2 / f
Specific Heat of Solids
Molar specific heat of most of the solids at constant volume is equal to 3R
Debye Temperature
Temperature at which the molar specific heat of a solid at constant volume becomes equal to 3R
Specific heat of water (Cv Value)
Cv = 9R
Mean Free Path
The average distance travelled by the molecule between two successive collisions
Mean free path factors
- Proportional to m
- Inversely proportional to density
- Inversely proportional to d^2
- Proportional to absolute temperature
- Inversely proportional to Pressure
Lambda expression (mean free path)
lambda = (kT) / (root 2 * pi * d^2 * P)
Calculate number of molecules in 2 * 10^-6 m^3 of a perfect gas at 300K and at a pressure of 0.01m of mercury. Mean KE of a molecule at 300K = 4* 10^-11 J and g = 9.8
P = h ro g = 13.6 * 98
1/2 Mv^2 = 3/2 PV = Total KE
Total KE / KE per molecule = 10^8
Higher specific heat capacity = (rate of change of temp)
Lower rate
Higher specific heat capacity means heat is used for not onloy in KE but also for translational, rotational, etc.
A box contains equal number of molecules of hydrogen and oxygen. If there is a fine hole in the box, which gas will leak rapidly?
Hydrogen
v proportional to 1/ root (M)
Volume at STP
22.4 * 10 ^-3 m^3
Molecular weight in SI
Molecular weight is in grams
Multiply by 10 ^ -3
Relation brtween velocity and temperature
v2 / v1 = root (T2 / T1)
The molecular KE of 1g of helium at 400K is?
KE = 3/2 RT
KE = 3/2 * 400 * 8.31 * 1/4
= 12.465 J
