Heat and Thermodynamics Numerical

Question 1.
An ideal diatomic gas is enclosed in an insulated chamber at temperature 300K. The chamber is closed by a freely movable massless piston, whose initial height from the base is 1m. Now the gas is heated such that its temperature becomes 400 K at constant pressure. Find the new height of the piston from the base.If the gas is compressed to initial position such that no exchange of heat takes place, find the final temperature of the gas.

Question 2.
A cylinder of mass 1 kg is given heat of 20000J at atmospheric pressure. If initially temperature of
cylinder is 200C, find
(a) final temperature of the cylinder.
(b) work done by the cylinder.
(c) change in internal energy of the cylinder.
(Given that Specific heat of cylinder= 400 J kg-1 0C–1, Coefficient of volume expansion
= 9 × 10-5 °C-1, Atmospheric pressure = 105 N/m2 and Density of cylinder = 9000 kg/m3)

Question 3..
A gas container has a walls that can bear maximum pressure of 2.2*106 pa.The container contains gas at 4*105 pa and 350K.If the container is steadily heated then neglecting change in volume .Calculate the temperature at which container will break?

Question 4.
A vessel of volume V=20L contains a mixture of hydrogen and helium at T=20&#176 C and Pressure P=2.0 atm. The mass of the mixture is equal to m=5g. Find the ratio of the mass of the hydrgen to that mass of the helium in the mixture Given R=.082 L.atm.mol-1.K-1

Question 5:
A gas is contained in a vertical, friction less piston cylinder device. The piston has a mass of 20 kg with a cross-sectional area of 20 cm2 and is pulled with a force of 100N. If the atmospheric pressure is 100 kPa, determine the pressure inside.

Question 6:
A piston-cylinder device has a ring to limit the expansion stroke. Initially the mass of Oxygen is 2 kg at 500 kPa, 30° C. Heat is now transferred until the piston touches the stop, at which point the volume is twice the original volume. More heat is transferred until the pressure inside also doubles. Determine the amount of heat transfer and the final temperature.

Question 7
A piston cylinder device contains 1 kg of oxygen at 150 kPa, 30°C. The cross-sectional area of the piston is 0.1 m2. Heat is now added causing gas to expand. When the volume reaches 0.2 m3, the piston reaches a linear spring with a spring constant of 120 kN/m. More heat is added until the piston rises another 25 cm. Determine the final pressure, temperature and the energy transfers.

Question 8
Is the energy of the system U an intensive or extensive variable?

Question 9
Suppose we have M+N systems prepared, and the first is in thermal equilibrium with the second, the second in thermal equilibrium with the third, etc., until system M+N- 1 is in thermal equilibrium with the M+Nth system. Is the first system in thermal equilibrium with the M+Nth?

Question 10
A gas is contained in a cylinder with a moveable piston on which a heavy block is placed. Suppose the region outside the chamber is evacuated and the total mass of the block and the movable piston is 102 kg. When 2140 J of heat flows into the gas, the internal energy of the gas increases by 1580 J. What is the distance s through which the piston rises?

Question 11
A resistance thermometer is such that resistance varies with temperature as
where T represent Temperature on Celsius scale And a,b,R0 are constants.R0 unit is ohm
Based on above data ,Find out the unit of a,b

Question 12.
A monotonic gas is taken through ABCDA cycle.

A->B Constant Pressure process

B->C Constant volume process

C->D Constant Pressure process

D->A Constant volume process

The PV coordinates of point A are (P,V)

The PV coordinates of point B are (P,3V)

The PV coordinates of point C are (3P,3V)

The PV coordinates of point D are (3P,V)

1 Find the total work done in the cycle

2 Change in Internal energy

Question 13
The pressure of the gas in constant volume gas thermometer are 80 cm,90cm and 100cm of mercury at the ice point,the steam point and in a heated wax bath resp.Find the temperature of the wax bath

Question 14
The heat capacity of a substance is found to be varies as temperature


A sample of mass m of that substance is heated from temperature T1 to T2. How much heat is required

Solution 5
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

Solution 6

Since O2 is a diatomic gas
Molecular mass=32 gm

Initial state
Initial volume of gas
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

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

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

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

Solution 7

Since O2 is a diatomic gas
Molecular mass=32 gm

Initial state
Initial volume of gas
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
=200 KJ

Work done 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 8
Doubling the system should double the energy, so U is an extensive variable.

Solution 9

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

Solution 10

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
s=.54 m

Solution 11

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

Solution 12
Work done in Process AB=P(3V-V) =2PV
Work done in Process BC=0 as volume is constant
Work done in Process CD=3P(V-3V) =-6PV
Work done in Process DA=0 as volume is constant S
o net work done=-4PV
Net heat =Net work done=-4V
 Change in Internal energy =0

Solution 13
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)/(Psteam-Pice) x 100

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

Solution 14
Now integrating with upper and lower limit as Tand T2
Q=mC0[2T2-2T1-a(T12-T22]/2 =mC0(T2-T1)(2+a(T2+T1))/2

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