3 Marks Question

Question 13 Marks

Should the internal energy of a system necessarily increase if heat is added to it?

Answer

Change in internal energy of a system, $\Delta\text{U}=\text{C}_\upsilon\Delta\text{T}$
Here,
C_v = Specific heat at constant volume
$\Delta\text{T}$ = Change in temperature.
If $\Delta\text{T}=0,$ then $\Delta\text{U}=0,$ i.e. in isothermal processes, where temperature remains constant, the internal energy doesn't change even on adding heat to the system.
Thus, the internal energy of a system should not necessarily increase if heat is added to it.

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Question 23 Marks

What should be the condition for the efficiency of a Carnot engine to be equal to 1?

Answer

Efficiency of Carnot engine, $\eta=\frac{\text{W}}{\text{Q}_1}=\frac{\text{Q}_1-\text{Q}_2}{\text{Q}_1}$
Here,
W = Work done by the engine
Q₁ = Heat abstracted from the source
Q₂ = Heat transferred to the sink
Thus, for
$\eta=1,\text{W}=\text{Q}_1\ \text{or}\ \text{Q}_2=0,$ the total heat which is abstracted from the source gets converted into work.

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Question 33 Marks

When we heat an object, it expands. Is work done by the object in this process ? Is heat given to the object equal to the increase in its internal energy?

Answer

When we heat an object, it expands, i.e. its volume increases.
Work done by the system, $\Delta\text{W}=\text{P}\Delta\text{V}$
Using the first law of thermodynamics, we get,
$\Delta\text{Q}=\Delta\text{U}+\Delta\text{W}$
Since the volume changes, $\Delta\text{W}$ has some non-zero positive value. Thus, heat given to the object is not equal to the increase in the internal energy of the system.

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Question 43 Marks

Calculate the change in internal energy of a gas kept in a rigid container when 100J of heat is supplied to it.

Answer

Q = 100J
We know, $\Delta\text{U}=\Delta\text{Q}-\Delta\text{W}$
Here since the container is rigid, $\Delta\text{V}=0,$
$\Delta \text{V}=\text{V}_\text{f}-\text{V}_\text{t}=0$
Hence the $\Delta\text{W}=\text{P}\Delta\text{V}=0,$
So, $\Delta\text{U}=\Delta\text{Q}=100\text{J}$

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Question 53 Marks

A cylinder containing a gas is lifted from the first floor to the second floor. What is the amount of work done on the gas? What is the amount of work done by the gas? Is the internal energy of the gas increased? Is the temperature of the gas increased?

Answer

As a cylinder is lifted from the first floor to the second floor, there is decrease in the atmospheric pressure on the gas and it expands. Therefore, some work is done by the gas on its surroundings. Work done on the gas is zero.
Work done by the gas, $\text{W}=\text{P}\Delta\text{V}$ (positive)
The increase in the internal energy and temperature of the system will depend on the types of the walls of the system (conducting or insulating).

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Question 63 Marks

Figure. shows three paths through which a gas can be taken from the state A to the state B. Calculate the work done by the gas in each of the three paths.

Answer

In path ACB,

W_AC + W_BC = 0 + pdv = 30 × 10³(25 - 10) × 10^-6 = 0.45J
In path AB, W_AB $=\frac{1}{2} \times (10 + 30) \times 10^3\times 15 × 10^{-6} = 0.30\text{J}$
In path ADB, W = W_AD + W_DB = 10 × 10³(25 - 10) × 10^-6 + 0 = 0.15J

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Question 73 Marks

When an object cools down, heat is withdrawn from it. Does the entropy of the object decrease in this process? If yes, is it a violation of the second law of thermodynamics stated in terms of increase in entropy?

Answer

When an object cools down, heat is withdrawn from it. Hence, the entropy of the object decreases. But the decrease in entropy leads to the transfer of energy to the surrounding. The second law is not violated here, which states that entropy of the universe always increases as the net entropy increases.
Here,
Net entropy = Entropy of object + Entropy of surrounding

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Question 83 Marks

When we stir a liquid vigorously, it becomes warm. Is it a reversible process?

Answer

When we stir a liquid vigorously, we do work on it due to which its temperature rises and it becomes warm. To bring back the liquid to the initial temperature, heat needs to be extracted from it. But it is an irreversible process that cannot be brought back to the initial state by stirring the liquid again in the opposite direction.

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Question 93 Marks

The final volume of a system is equal to the initial volume in a certain process. Is the work done by the system necessarily zero? Is it necessarily non-zero.

Answer

Work done by the system is neither necessarily zero nor necessarily non-zero.
If in a certain process, the pressure P stays constant, then,
$\Delta\text{W}=\text{P}\Delta\text{V}$
$\Rightarrow\text{W}=\text{P}(\text{V}_2-\text{V}_1)$
As $\text{V}_2=\text{V}_1$
$\Rightarrow\text{W}=0$
(Initial volume, V₁= Final volume, V₂)
Hence, it is an isobaric process.
Even if P = P(V), net work done will be zero if V₂= V₁. In this case, work done is zero.
If the system goes through a cyclic process, then initial volume gets equal to the final volume after one cycle. But work done by the gas is non-zero.

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Question 103 Marks

When a system is taken through the process abc shown in figure. 80J of heat is absorbed by the system and 30J of work is done by it. If the system does 10J of work during the process adc, how much heat flows into it during the process?

Answer

$\Delta\text{Q}=\Delta\text{U}+\Delta\text{W}$

In abc $\Delta\text{Q}=80\text{J},\ \Delta\text{W}=30\text{J}$

So, $\Delta\text{U}=(80-30)\text{J}=50\text{J}$

Now in adc, $\Delta\text{W}=10\text{J}$

So, $\Delta\text{Q}=10+50=60\text{J}\ [\therefore\Delta\text{U}=50\text{J}]$

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Question 113 Marks

A substance is taken through the process abc as shown in figure. If the internal energy of the substance increases by 5000J and a heat of 2625cal is given to the system, calculate the value of J.

Answer

Now, $\Delta\text{Q}=(2625\times\text{J)J}$

$\Delta\text{U}=5000\text{J}$

From Graph $\Delta\text{W} = 200 \times 10^3 \times 0.03 = 6000\text{J.}$

Now, $\Delta\text{Q}=\Delta\text{W}+\Delta\text{U}$

$\Rightarrow2625\text{J}=6000+5000\text{J}$

$\text{J}=\frac{11000}{2625}=4.19\text{J/cal}$

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