## Pages

### 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
CP=7R/2,CV=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=nCP(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=nCV(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
CP=7R/2,CV=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=nCP(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=nCV(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