Class 12-science H C VERMA Solutions Physics Chapter 5 - Specific Heat Capacities of Gases
Specific Heat Capacities of Gases Exercise 77
Solution 1
Mechanical energy is converted into internal energy. Thus,
Solution 2
We know,
dQ=dU+dW
dQ=dU (As dW=0)
Q=8.6cal
Or Q=36.12J
Solution 3
We know,
dW=PdV
dW=300J
Also, dW=nRdt
And 
dQ=1050J
Solution 4
We know,
So
Also,
Solution 5
a) We know,
dQ=dU+dW
dU=dQ-dU
=6RdT
dU=2490J
b) dU=dQ
dU=2490J
c) Adiabatically 
dU=2490J
Specific Heat Capacities of Gases Exercise 78
Solution 6
We know,
PV = nRT
Also,
Also,
=51.18
=2.08cal
Solution 7
a) We know,
dQ=dU+dW
dU=dQ-dW
=dQ-PdV
dU=30J
b) For monoatomic gas
n=0.008
c) We know,
And 
=12.5+8.3=20.3
d) From equation 1
Solution 8
We know,
Also,
Q=3nRT
Or
Solution 9
As given:
P=kV
RdT=2kVdV
We know,
dQ=dU+dW
Solution 10
As given
Solution 11
For 
ideal gas:
For 
ideal gas:
As given 
i.e. 
When we mix the gas
Also,
Taking ratio of equation 1 and 2
Solution 12
We know,
Also,
Ratio 
Solution 13
a) We know,
b) We know,
=1250J
Also, dQ=dU+dW [[For bc process, dW=0]
c) Heat liberated, 
=750J
d) Also, Heat libeated, (for cd process)
=2500J
Solution 14
a) In case of ab , 
is constant
Thus, 
In case of bc, Pis constant
Thus 
b) Work done=Area under graph
c) We know,
Q=14.9J
And also,
Q=24.9J
d) We know,
=(24.9+14.9)-1 [from1 and 2]
Solution 15
We know
Solution 16
a) We know
b) We know,
Solution 17
a) 
b)
c) Workdone is given as:
W=21J
Solution 18
A diabetic process
Thus,
Solution 19
a) We know
Also,
b) When the container is slowly compressed then also heat transferred is 0 because walls are adiabatic and thus
Pressure p=800kPa
And temperature T=600K
Solution 20
a) First slowly compressed ∴Isothermal compression.
So,
Now, suddenly compressed ∴Adiabatic compression.
b) First gas is
compressed suddenly
Adiabatic compressin
Now, gas is compressed slowly ∴ Isothermal compression.
Solution 21
a) Isothermal condition
Now, adiabatic condition. Thus,
b) Adiabatic condition:
Now, isothermal condition,
Solution 22
a) We know,
PV=nRT
n=0.009moles
b) We know,
c) Adiabatic process ∴
Or
d) We know,
W=-33J
e) We know, internal energy
W=33J
Specific Heat Capacities of Gases Exercise 79
Solution 23
A is isothermal, Thus.
B is adiabatic. Thus,
C is isobaric process. Thus,
And 
As given
Ratio is given as:
Solution 24
Work done is given as:
And
As work done is equal for both cases, Thus,
Also we know,
Substituting above equation in 3 we get
Solution 25
a) Adiabatic process, Thus,
b) We know,
And also
Now
c) We know,
dQ=dU+dW
dU=-dW [As dQ=0]
dU=+82J
d) from equation 1
e) Isobaric process as pressure is constant. Thus,
f) Here, 
For isothermal condition,
g) Net work done, 
=-82-41.4+103
=-20.4J
Solution 26
In this, adiabatic process is going on. So
So, 
Solution 27
a) We know,
PV=nRT
n=0.008
b) We know,
dQ=dU+dW
dQ=dU (dW=0,as no change in volume)
For A,
And 
For B,
And 
Distance moved by
mercury =
=25-12.5
Solution 28
Adiabatic process, Thus.
For 
,
For 
, 
Where m is mass of H2
Equating 1 and 2 we get,
m=0.029
m=0.03g
Solution 29
a) Temperature remains constant as A is diathermic
Thus, 
And 
Now as B is adiabatic,
Solving we get
Also,
b) Here, temperature remains constant because value is open.
And
A is diathermic thus, T and V are constant and so pressure also remains constant.
B is adiabatic. Thus,
Or 
Solution 30
a) Adiabatic process,
And
When equilibrium is reached then
Solving we get
And
b) Process is adiabatic so no heat is transferred to left part.
Q=0
c) From equation 1
And from equation 3
Thus, equation 5 becomes:
Solution 31
Process is adiabatic. Thus,
Also,
V=447m/s
Specific Heat Capacities of Gases Exercise 80
Solution 32
We know,
Spedd of sound, 
Also,
=18.0J/mol-x
Also,
=26.3J/mol-K
Solution 33
We know,
Also, speed od sound V=
V=960m/s
Solution 34
In kundt's tube, node separation is given as:
Speed of sound 
=360m/s
Also,
Thus,
Now, we know,
Also,
Solution 35
Speed of sound 
=330m/s
Also,
{PV=nRT
}
So, 
We know,
And
