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=nC^VΔT
=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 P_ice= 80 cm of Hg
Pressure at the steam point P_steam= 90 cm of Hg
Pressure at the wax bath P_wax= 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