Question-1.Let A & B are two sample of ideal gases of equal mole .let T be the temperature of both the gas Let E_A and E_B are there total energy respectively .Let M_A and M_B are these respective molecular mass .which of these is true
a,E_A > E_B
b,E_A < E_B
c,E_A =E_B
d,none of these

Solution:1
E_A = 3/2 nRT
E_B = 3/2 nRT
;E_A =E_B


Question-2. The velocities of the molecules are v, 2v, 3v, 4v & 5v. The rms speed will be
a,11v
b,v(11)^1/2
c, v
d, 3.3v

Solution:2
Vrms= (∑ V² / N)^1/2
= [(V² + 4V² + 9V² + 16V² + 25V²)/5]^1/2
=v(11)^1/2


Question-3. What is true of Isothermal Process
a, ΔT >0
b, ΔU=0
c ΔQ=ΔW
d PV=constants

Solution-3:

In an Isothermal Process
Temperature remains constant ΔT =0
Since Internal energy depends on the temperature
ΔU=0

From first law of Thermodynamics
ΔU=ΔQ-ΔW
Since ΔU=0
ΔQ=ΔW

Also PV=nRT
As T is constant
PV= constant



Question-.4 Two absolute scales A and B have triple points of water defined as 200A and 350A. what is the relation between T_A and T_B



Solution-4

Given that on absolute scale
Triple point of water on scale A = 200 A
Triple point of water on scale B = 350 B
Also, triple point of water on Kelvin scale = 273.16 K
Now temperature on scale A and on scale B is equivalent to 273.16 K on absolute temperature scale.
Thus, value of one degree on absolute scale A = (273.16/200) K
Or,
Value of temperature T_A on absolute scale A = (273.16XT_A)/200
Similarly value of temperature T_B on absolute scale B = (273.16XT_B)/350
Since T_A and T_B represent the same temperature
273.16×T_A/200 = 273.16×T_B/350
Or, T_A = 200T_B/350 = 4T_B/ 7

Question 5:A gas is contained in a cylinder with a moveable piston on which a heavy block is placed. Suppose the region outside the chamber is evacuated and the total mass of the block and the movable piston is 102 kg. When 2140 J of heat flows into the gas, the internal energy of the gas increases by 1580 J. What is the distance s through which the piston rises?

Solution:5

Total heat supplied =Workdone + Change in internal energy

So work done=2140-1580=560 J

Let s be the distance moved then
the workdone is given by =Fs
Fs=560
s=560/F
=560/102*10
s=.54 m


Question 6:
The surface of a body has a emissivity of .55 and area of 1.5 m²
Find out the following
a. What rate of heat is radiated from the body if the temperature is 50°C
b. At what rate is radiation absorbed by the radiater when the surrounding temperature is 22°C
c What is the net rate of radiation from the body
Given σ=5.67 *10^-8

Solutions 6
a) Rate of radiation radiated=eσAT_b⁴=(.55)(5.67 *10^-8)(1.5)(323)⁴=509W
b) Rate of radiation absorbed=eσAT_s⁴=(.55)(5.67 *10^-8)(1.5)(295)⁴=354W
c)Net =Rate of radiation radiated-Rate of radiation absorbed=155 W

Question-7: At 27°C,two moles of an ideal monoatomic gas occupy a volume V.The gas is adiabatically expanded to a volume 2V.
Calculate the ratio of final pressure to the intial pressure
Calculate the final temperature
Change in internal energy
Calculate the molar specific heat capacity of the process

Solution-7
Given
n=2 T=27°C=300 K ,V₁=V,V₂=2V

Now PV^y=constant
P₁V₁^y=P₂V₂^y
P₂/P₂=(V₁/V₂)^y
=.5^5/3

ALso
T₁V₁^y-1=T₂V₂^y-1
or T₂=300/2^5/3=189K

Change in internal energy=nC_vΔT
For monoatomic gas C_v=3R/2
Substituting all the values
Change in internal energy==-2764J

As in adiabatic process ΔQ=0,molar specific heat capacity=0


Question-8
An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C.It absorbs 6*10² cal of heat at the higher temperature.Calculate the amount of heat supplied to the engine from the source in each cycle

Solutions-8:
T₁=227°C =500K
T₂=127°C =400K

Efficiency of the carnot cycle is given by
=1-(T₂/T₁)=1/5

Now also efficency =Heat supplied from source/Heat absorbed at high temperature
so Heat supplied from source=6*10²*(1/5)==1.2*10²cal



Question-9:
An ideal gas A is there.Intial temperature is 27°C.The temperature of the gas is increased to 927°C.Find the ratio of final V_rms to the initial V_rms

Solution-9:

V_rms=√3RT/M

So it is proportional to Temperature
Now
T₁=27°C =300K
T₂=927°C =1200K

So intial V_rms=k√300
Final V_rms=k√1200

ratio of final to intial=2:1

Question-10:
A circular hole of diameter 2.00 cm is made in an aluminium plate at 0 ° C .what will be the diameter at 100° C?
α for aluminium = 2.3 * 10^-3 / ° C

Solutions-10. Diameter of circular hole in aluminium plate at 0° C=2.0 cm
With increase in temperature from 0° C to 100° C diameter of ring increases using

L=L₀(1+αΔT)

where L₀=2.0 cm
α = .3 * 10^-3 / ° C
ΔT=(100 -0)=100 ° C
we can find diameter at 100° C
L=2(1+2.3*10^-5*100° C)
      =2.0046 cm