Solution of Jan 25 Problems

Solution 1:
Total forces on the piston
Weight of the piston acting downward=200N
Force due to Atmospheric pressure downward =100*10³*20*10^-4 N=200N
Let P be the Pressure of the gas
Then Force acting due to pressure of gas on the piston upward=PA
Force with the piston is being pulled upward=100N

Total Upward force=Total Downward force
PA + 100=200 +200
20*10^-4*P=300
P=150kPa

Solution 2:

Since O₂ is a diaatomic gas
C_P=7R/2,C_V=5R/2
Molecular mass=32 gm

Intial state
Intial volume of gas
V=nRT/P
n=2*10³/32
T=303K
R=8.3
P=500*10³ N/m²
so V=.314 m³

Second state when At top volume becomes double

Volume=2*.314 =.628m³
Pressure=500*10³ N/m² remains same

So as per ideal gas equation
T=606K

Workdone by the gas=PV=500*10³*.314=157*10³ J
Heat transfered=nC_P(T₂ -T₁)
=2*10³ *7*8.3*303/32*2
=550*10³ J

Third state when Pressure becomes double
P=1000*10³ N/m²
V=.628m³

As per ideal gas equation
Temperature=1212K
Heat transfered=nC_V(T₃ -T₂)
=2*10³ *5*8.3*606/32*2
=785 kJ

So total heat tranfered=550+785=1335 kJ
Final Temperature=1216 K

Solution 3:
Since O₂ is a diaatomic gas
C_P=7R/2,C_V=5R/2
Molecular mass=32 gm

Intial state
Intial volume of gas
V=nRT/P
n=1*10³/32
T=303K
R=8.3
P₀=500*10³ N/m²
so V=.157 m³

Second state ( when piston reaches the spring)
v=.2 m³
P=500*10³ N/m²
As per ideal gas equation
Temperature becomes=385.5 K


So heat tranfer till that point=nC_P(T₂ -T₁)
=1*10³ *7*8.3*82.5/32*2
=74.8 KJ

Third state ( when it compresses the spring)V=.2+0.1*0.25=0.225 m³
P=500*10³ + kx/A
=500*10³ + 120*10³*25*10^-2/.1
=800*10³ N/m²

As per ideal equation
Temperature becomes=694K
Change in Internal energy=nC_V(T₃ -T₂)=5nR(T₃ -T₂)/2
=5*1*10³*8.3*308.5/32*2
=200 KJ

Workdone by the gas =P₀Ax + kx²/2
=500*10³*.1*.25 + 120*10³*.25*.25/2
=16.25 KJ

Total heat supplied in this Process=200+16.25=216.25KJ

So net Heat transfer=291.05 KJ

Solution 4:
Doubling the system should double the energy, so U is an extensive variable.


Solution:5
By repeated application of the Zeroth Law, we can state that all M+N systems are in thermal equilibrium with each other.

Solution:6

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

Solution:7
As per dimension analysis Unit on both sides should be equal
Now since R & R₀ both unit are same
Quantity 1+aT+bT⁵ should be dimension less
so at should be dimension less
so a unit is C^-1
similarly bt⁵ should be dimensionless
so b unit is C^-5