IIT JAM Physics MCQ Test 7 - Physics MCQ

Mean molecular energy 

Since the r..m.s. velocity  we have  
Here T₁ =273K and  So. we have, 

For one mole of an ideal gas,
PV = RT and pt = k (constant)
∴ 
So 
∴ Work done, when temperature changes from T to (T + dT),

NOW, dQ-dU + dW
CcTT - C_vdT +2R dT
∴ 
Hence, degree of freedom 

We have    4813 J.k mol^-1 K^-1

Heat lost by Ni = heat gained by water + Calorimeter 
We know heat capacity of Ni. CNi = 0.106 kcal/⁰C 

So. 0.250 (0.106) (120⁰ - t ) = 
3.18 - 0 .071 = 0.2201 - 2.20  
0.247t= 5.38

The cyclic process 1 is clockwise and the process 2 is anti clockwise. Therefore W₁, will be positive and W₂ will be negative area 2 > area 1, Hence the net work will be negative.
Hence the correct answer will be (b)

The change in internal energy dU is independent of the path. i.e.. if initial and final states of change are same then dU will be same.


Here, in both cases initial states and final states are respectively same. Hence. dU will be same in both ⇒ 

Q₁ =1000 calories
Q₂ =400 calories
T₁ = 500 k

Slope of adiabatic curve = γ x slope of isothermal curve.

This is shown by every cuive given in question . Hence . all curves represent the ideal gas.

The energy density of a photon gas is 
U = σT4
Thus the energy is

Now use the first law of thermodynamics
dE = dQ - dW
dQ = dE + dW = dE + pdV

N₂(g) + 3H₂ → 2NH₃(g)
∆s = 2 – 4 = – 2 < 0
So entropy is negative.
Hence option (c) is the answer.

dT = 
dP: represents the fall in temperature which is always negative.
Now dT is negative if αT > 1
So, cooling is produced
dT is positive if  or heating is produced.

Magnetic moment for spin 1/2 system
 ...(i)
∴ Partition function.     ...(ii)
from (1) and (2). we get

thus 
For N spins 
Helmholtz free energy, 

Partition function

for composite system,
  Given Z₁ = Z₂
So, Z = Z^2N

The Clapeyronss equation describing the rate of change of vapour pressure with temperature is

Or 
Here L = 540 cal. = 540 - 4.2 - 10⁷ ergs., T = 100 + 273 = 373 K
dT = ( 103 - 100 ) = 3°, V₂ = 1670 c.c./gm., V₁= 1 c.c./gm. 
∴ Change in vapour pressure


= 0.17091 atmospheres.

From eq. we know that

Here 

Hence 
= 7.5 x 10^-7 m
The number of collisions per unit distance is the inverse of the mean free path.

The average speed of a molecule obeying maxwellian distribution is given by

Collision frequency is

The mean free time, or the average time between collisions is the inverse of the collision frequency and in this case will be equal to

According to 1st law of thermodynamics,
ΔQ = ΔU+ΔW
So for the vessel for which internal energy (and hence, temperature) remains constant


i.e., 
and for the vessel for which volume is kept constt., 
 
i.e.,  
According to given problem 


i.e., 
so 
Now when the free mixing of gases is allowed
u₁+u₂ = u
 With 
Here  and 
so 1 x 300 + 1 x 438.6 = 2T, i.e., T = 369.3 K
Further for the mixture from PV = μRT with V = V + 2V = 3V and μ = μ₁ + μ₂ = 2.
we have 

(a) The work done by the vaporizing water is 
W = p(V₂ - V₁)
= (1.013 x 10⁵ Pa) (1671 x 10^-6 m3 ^-1 x 10^-6m³)
= 169 J
(b) The heat added to the water to vaporize it is 
Q = mL_v = (10^-3 kg) (2.256 x 10⁶ J/kg) = 2256 J
(c) From the first law of thermodynamics, the change in internal energy is 
ΔU= Q — W = 2256 J - 169 J = 2087 J

(P,V,T,S) are four thermodynamic variables but only two of them are idependent variables while the other two may in be expressed as their function.
(i) Representing the pressure P as a function of twp omde[emdemt varoan;es S amd T i.e.
P = P (S,T)
∴ 
If the process is isobaric i.e., change takes place at constant pressure, dP = 0. 
Then

or, 
or, 
(ii) Representing P as a function of T and V i.e.
P = P (T,V)
∴ 
But for isobaric process dP = 0. Hence

or 
or 

The atoms or molecules having an even number of fundamental particles have integral spin value and are called Bosons. Similarly, the atoms or molecules having odd number of fundamental particles have half integral spin value and are called Fermions. Accordingly, the classification of particles is given below:
Proton, Neutron, electron and positron are fundamental particles having half spin value. Hence they all are Fermions.
Photon has spin 1 and hence it is Boson

Hydrogen-molecule = 
Positron 
Lithium ion 

If  are the average speeds of hydrogen and nitrogen molecules at temperatures  T₁K and T₂K respectively, we have
 ...(i)
 ...(ii)
where m₁ and m₂ are the masses of hydrogen and nitrogen molecules respectively.
Dividing eq. (1) and (2), we get
 ...(iii)
Given when  
and 
Substituting these values in (3), we get

or 
The most probable speed of nitrogen at temperature T is given by

Here k = 1.38 * 10^-23 joule per degree, T = 22 K 
and m = 28 a.m.u. =
∴ 
= 1.14 * 10²m/s.

K.E. of a molecule of mass m and mean square speed 

But 
∴ 
Escape velocity.  where R is the radius of the earth. 
Given
∴ K.E. of a molecule =  
∴ 
Or 
Temperature for hydrogen


Temperature for oxvaen.


= 16 x 10⁴K.

We have PV = nkT
or 
As total number of air molecules in the bulbs would remain constant, we have
n₁ + n₂ = n₁ + n₂
As temperature of second bulb is unchanged, we have

i.e. 
or 
This gives new pressure

Here P₀ = 76 cm o f Hg,

∴ 
 = 83.75 cm of Hg
 

Let m and P be the initial mass and pressure of the gas inside the vessel. Therefore,
PV = (m/M) RT ...(1)
where M is the molecular weight of the gas in the vessel.
After a part of the gas is released , we have
 ...(ii)
where , m is the mass of the remaining gas in the vessel.
Hence, mass of the gas released is equal to (subtracting (2) from (1)

Now, under normal conditions (P₀ = 1 atm. T = 273K), density of the gas is given  to be p . Therefore, we find.

or, M / RT = P/ PO
Thus, 
Now here  V = 30 * 10^-3 m³

and, P_o =1atm.
Therefore, 

Initially when the gas is heated, pressure remains constt.
SO  ...(i)
Now when the piston is clamped, volume becomes constt. 
So heat withdrawn ..(ii)
So the difference between heat added and removed,

But here 
SO  
i.e., Ans.

By conservation of energy Q₁ = W + Q₂ + Q₃
i.e., Q₂ + Q₃ = Q₁ - W = 1200 - 200 = 1000 (1)
And as change in entropy in a reversible-process is zero
 i.e. 
i.e., 2Q₂ + 3Q₃ = 1800 (2)
Solving equation (1) and (2) for Q₂ and Q₃, we get 
Q₂ = 1200 J and Q₃ = -200 J
i.e.. the reservoir at temperature T₂ absorbs 1200 J of heat while the reservoir at temperature T₃ lose 200 J of heat. Hence their sum will be 1400 J

This process involves irreversible heat flow because of the temperature differences.
Let final temp is 


So. the final temperature is 50⁰C = 323 K. The entropy change of the hot water is


The entropy change of the cold water is

The total entropy change of the system is

An irreversible heat flow in an isolated system is accompanied by an increase in entropy. We could have reached the same end state by simply mixing the two quantities of water. This. too. is an irreversible process; because the entropy depends only on the state of the system . the total entropy change would be the same, 102J/K.

The critical temperature is given by

given a = 0.00874, b = 0.0023 ..(1)
and 
Substituting these values in equation (1). we get

= 305.5⁰C.

Given dT = 1⁰C . V₂ - V₁ = -0.091 c.c.
L = 80 cal. = 80 * 4.2 * 10⁷ ergs, T = 0 + 273 = 273 K 
Applying Clapeyrons latent heat equation

or  dynes/cm².

= 135.2 Atmos.
Hence to lower the melting of ice by 1°C, the pressure must be raised by 135.2 atmospheres.
Now melting point of ice at atmospheric pressure is 0°C. Hence pressure required to make ice freeze at - 1 °C 
= 135.2 + 1 = 136.2 atmospheres

As the expansion is against zero pressure, so no work is done. Now, for given mass of gas,
PV = P₁V₁
= P₁ x 4V
∴ P₁ = P/4 is the pressure of gas after expansion.
In the second process, compression is adiabatic, so 

Change in internal energy = work done in adiabatic compression.

= 8P₁V = 2PV
= 2 * 2 * 10⁵ * 10^-3 = 400J
Further, 
∴ 
= 12.05
∴ Final temperature = T₁ + (T₂ - T₂)
= 300 + 12.05 = 312.05 K