Solution for Jan 28 Problems

Solution 1:
a. Let P be the Pressure of the gas.
Then
Pressure of gas=weight of the Piston/A + Atmospheric pressure
=Mg/A + P N/m2

b.Pressure remain constant during the heating process
Intial state of Gas
P1=Mg/A + P
T1=T

Now we know that
PV=nRT


V1 = nRT1
P1

V1 = nRT
Mg/A + P



Final State
P2=Mg/A + P
T2=2T

Now we know that
PV=nRT


V2 = nRT2
P2

V2 = 2nRT
Mg/A + P



Workdone by gas=P(V2-V1)

Substituting all the values we get
Workdone=nRT

Change in Internal Energy=nCVΔT
=n*3R/2*T
=3nRT/2

Now from first law of thermodynamics
Heat supplied=Change in Internal Energy + Workdone by gas
=5nRT/2


Solution 2:
Given that
Pressure at the ice point Pice= 80 cm of Hg
Pressure at the steam point Psteam= 90 cm of Hg
Pressure at the wax bath Pwax= 100 cm of Hg

The temperature of the wax bath measured by the thermometer is


T = (Pwax-Pice)*100
Psteam-Pice



T = (100 - 80)X100/(90-80)
= 20X100/10
=200° C


Solution 3:

dQ=mCdT
Now integrating with upper and lower limit as T1 and T2
Q=∫mC0(1+aT)dT
Q=mC0[T+aT2/2]

Q = mC0[2T2-2T1-a(T12-T22]
2
Q = mC0(T2-T1)(2+a(T2+T1))
2


Solution 4:

Workdone in Process AB=P(3V-V)
=2PV
Workdone in Process BC=0 as volume is constant
Workdone in Process CD=3P(V-3V)
=-6PV
Workdone in Process DA=0 as volume is constant

So net workdone=-4PV

Net heat =Net workdone=-4V
Change in Internal energy =0

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

Solutions for Jan 18 problems

Solution. 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 = 2.3 * 10-3 / ° C
ΔT=(100 -0)=100 ° C
we can find diameter at 100° C
L=2(1+2.3*10-3*100)
=2.46 cm

Popular Posts