Question 1 1 / -0
During the adiabatic expansion of 2 moles of a gas, the internal energy was found to have decreased by 100 J. The work done by the gas in this process is
Solution
dU = -100 J
Question 2 1 / -0
For a diatomic gas, change in internal energy for a unit change in temperature for constant pressure and constant volume is U1 and U2 , respectively. What is the ratio of U1 to U2 ?
Solution
In case of the diatomic gas the internal energy is given by U =
nRT
As the temperature of the both gases is same.
Hence U
1 : U
2 :: 1 : 1
Question 3 1 / -0
The work of 146 kJ is performed in order to compress one kilo mole of a gas adiabatically, and in this process, the temperature of the gas increases by 7℃. The gas is-1 K-1 )
Question 4 1 / -0
The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is
Solution
Q
1 = T
0 S
0 +
T
0 S
0 =
T
0 S
0 Q
2 = T
0 (2S
0 - S
0 ) = T
0 S
0 and Q
3 = 0
=
Question 5 1 / -0
A polyatomic gas (
= 4/3) at pressure P is compressed to (1/8)
th of its initial volume adiabatically. The pressure will change to
Solution
The adiabatic equation of an ideal gas is
Question 6 1 / -0
Calculate the temperature, in K, at which a perfect black body radiates at the rate of 5.67 watts cm-2 , σ = 5.67 × 10-8 watts m-2 K-4 .
Solution
Rate of energy dissipated per unit area per second is given by:
R = 5.67 watts cm
-2 = 5.67 × 10
4 watts m
-2
Question 7 1 / -0
Find the actual mechanical advantage of a machine, which has a velocity ratio of 3.2 and efficiency 75%.
Question 8 1 / -0
A heat engine takes in 900 J of heat from a high temperature reservoir and produces 300 J of work in each cycle. What is its efficiency?
Question 9 1 / -0
If a system changes from state (P
1 , V
1 ) to (P
2 , V
2 ) as shown below, the work done by the system is
Solution
Work done = Area of graph ACD
Work done =
Question 10 1 / -0
In the given figure, when a system is taken from state i to state f along the path iaf, it is found that heat Q absorbed by the system is 50 cal and work W done by the system is 20 cal. Along the path ibf, Q given to system is 36 cal. What is the value of work done by system along the path ibf?
Solution
According to the first law of thermodynamics,
Question 11 1 / -0
Which of the following statements are incorrect regarding the first law of thermodynamics?
Solution
Statements A and D are wrong.
Question 12 1 / -0
N moles of a monoatomic gas are carried round the reversible rectangular cycle ABCDA as shown in the diagram. The temperature at A is T
0 . What is the approximate thermodynamic efficiency of the cycle?
Solution
Q
AB = NC
v . T
0 Q
BC = NC
p . 2T
0 Q
CD = - NC
v . 2T
0 Q
DA = - NC
p . T
0 Heat taken = Q
1 = Q
AB + Q
BC = NT
0 (C
v + 2C
p )
Heat taken = Q
2 = Q
CD + Q
DA = NT
0 (2C
v + C
p )
Question 13 1 / -0
Directions:
Question 14 1 / -0
A Carnot engine having an efficiency of
= 1/10 as heat engine is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is
Solution
Coefficient of the performance
, where Q
2 is heat taken out at low temperature
Relation between the coefficient of performance (β) and the efficiency of the heat engine (η) is
Hence
Question 15 1 / -0
An ideal gas heat engine operates in a Carnot's cycle between 227ºC and 127ºC. It absorbs 6 x 104 J heat at high temperature. The amount of heat converted into work is
Question 16 1 / -0
An ideal gas, after going through a series of four thermodynamic states in order, reaches the initial state again (cyclic process). The amounts (Q) and work (W) involved in these states are1 = 6,000 J, Q2 = -5,500 J, Q3 = -3,000 J and Q4 = 3,500 J1 = 2,500 J, W2 = -1,000 J, W3 = -1,200 J and W4 = xJ
Solution
According to the given problem:
ΔQ = Q
1 + Q
2 + Q
3 + Q
4 = 6000 – 5500 – 3000 + 3500
ΔQ = 9500 – 8500 = 1000 J
ΔW = W
1 + W
2 + W
3 + W
4 = 2500 – 1000 – 1200 + x
= 300 + x
And as for cyclic process,
U
F = U
I , ΔU = U
F – U
I = 0
So, from first law of thermodynamics,
ΔQ = ΔU = ΔW
We have,
1000 = (300 + x) + 0
i.e., x = 1000 – 300 = 700 J
As efficiency of a cycle is defined as:
η =
η =
= 0.105
η = 10.5%
Question 17 1 / -0
One gram of water at a constant pressure of 1.01 x 105 Pa is converted into steam without any change of temperature. The volume of 1 gm of steam is 1671 cc and the latent heat of evaporation is 540 cal. The change in internal energy due to evaporation of 1 gm of water is
Solution
dW = P
= 1.01
=
= 40 cal. nearly.
Question 18 1 / -0
Consider the following statements (A) and (B). Read the statements and mark the correct answer.
Solution
Both statements are correct.
Question 19 1 / -0
Directions: Choose the correct option that suits Assertion and Reason from the given below :Assertion : The Carnot cycle is useful in understanding the performance of heat engines.Reason : The Carnot cycle provides a way of determining the maximum possible efficiency achievable with reservoirs of given temperatures.
Solution
Both statements are correct as from Carnot cycle, we can conclude that the efficiency of the heat engine is always less than 1.
Question 20 1 / -0
A Carnot engine absorbs 1000 calorie of heat from a source and rejects 800 calorie of heat to the sink per cycle.What is its thermal efficiency?
Solution
Efficiency of the heat engine
%
Question 21 1 / -0
An ideal gas heat engine operates in a Carnot cycle between 2270 C and 1270 C. It absorbs 6 kcal at the higher temperature. The amount of heat (in kcal) converted into work is equal to:
Solution
Here
T
1 = 127°C = 400 K, T
2 = 227°C = 500 K
Work done = efficiency xInput energy
W = 0.2 x 6= 1.2kcal
Question 22 1 / -0
A Carnot engine works as a refrigerator between 250 K and 300 K. If it receives 750 calories of heat from the reservoir at the lower temperature, then the amount of heat rejected at the higher temperature is
Question 23 1 / -0
A Carnot engine operates with a source at 500 K and a sink at 375 K. The engine consumes 600 kcal of heat per cycle. The heat rejected to the sink per cycle is
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