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### 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