Heat and thermodynamics solved questions

Question-1.Let A & B are two sample of ideal gases of equal mole .let T be the temperature of both the gas Let EA and EB are there total energy respectively .Let MA and MB are these respective molecular mass .which of these is true


a,EA > EB

b,EA < EB

c,EA =EB

d,none of these
Solution:1

EA = 3/2 nRT

EB = 3/2 nRT

;EA =EB

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 TA and TB

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 TA on absolute scale A = (273.16XTA)/200
Similarly value of temperature TB on absolute scale B = (273.16XTB)/350
Since TA and TB represent the same temperature
273.16×TA/200 = 273.16×TB/350
Or, TA = 200TB/350 = 4TB/ 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