Kinetic Theory of Gases and ldeal Gas | Physics JEE previous year questions with Solutions

Q 1 :

Water of mass m gram is slowly heated to increase the temperature from $T_{1}$ to $T_{2}$ . The change in entropy of the water, given specific heat of water is $1 {Jkg}^{- 1} K^{- 1}$ , is [2025]

zero
$m (T_{2} - T_{1})$
$m \ln (\frac{T_{1}}{T_{2}})$
$m \ln (\frac{T_{2}}{T_{1}})$

1

2

3

4

(4)

dQ = msdT

$d S = \frac{d Q}{T} = \frac{m s d T}{T}$

$△ S = \int \frac{m s d T}{T} = m s \ln \frac{T_{f}}{T_{i}}$

$△ S = m \ln \frac{T_{2}}{T_{1}} as s = 1$

Q 2 :

A Carnot engine (E) is working between two temperatures 473 K and 273 K. In a new system two engines - engine $E_{1}$ works between 473 K to 373 K and engine $E_{2}$ works between 373 K to 273 K. If $η_{12}$ , $η_{1}$ and $η_{2}$ are the efficiencies of the engines E, $E_{1}$ and $E_{2}$ , respectively, then [2025]

$η_{12} < η_{1} + η_{2}$
$η_{12} = η_{1} η_{2}$
$η_{12} = η_{1} + η_{2}$
$η_{12} \geq η_{1} + η_{2}$

1

2

3

4

(1)

Efficiencies of a Carnot engine $η = 1 - \frac{T_{sink}}{T_{source}}$

$\Rightarrow η_{1} = 1 - \frac{373 K}{473 K} = \frac{100}{473}$

$η_{2} = 1 - \frac{273 K}{373 K} = \frac{100}{373}$

$η_{12} = 1 - \frac{273 K}{473 K} = \frac{200}{473}$

$η_{12} - η_{1} = \frac{200}{473} - \frac{100}{473} = \frac{100}{473} < \frac{100}{373}$

$\Rightarrow η_{12} - η_{1} < η_{2}$

or $η_{12} < η_{1} + η_{2}$

Q 3 :

A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of the sink same, the new temperature of the source will be [2023]

900 K
300 K
1000 K
360 K

1

2

3

4

(3)

$T_{L} same$

$η = 1 - \frac{T_{L}}{T_{H}}$

$\frac{50}{100} = 1 - \frac{T_{L}}{600}$

$\frac{70}{100} = 1 - \frac{T_{L}}{T_{H}}$

$\frac{T_{L}}{600} = \frac{1}{2}, \frac{T_{L}}{T_{H}} = \frac{3}{10}$

$T_{H} = 1000 K$

Q 4 :

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. [2023]

Assertion A: Efficiency of a reversible heat engine will be highest at –273°C temperature of cold reservoir.

Reason R: The efficiency of Carnot’s engine depends not only on temperature of cold reservoir but it depends on the temperature of hot reservoir too and is given as $η = (1 - \frac{T_{2}}{T_{1}})$ .

In the light of the above statements, choose the correct answer from the options given below:

A is true but R is false
Both A and R are true but R is NOT the correct explanation of A
A is false but R is true
Both A and R are true and R is the correct explanation of A

1

2

3

4

(4)

$η = 1 - \frac{T_{cold}}{T_{hot}}$

$T_{cold} = 0 K \Rightarrow η = 1 (max)$

$Both A and R are true and R is the correct explanation of A.$

Q 5 :

A Carnot engine operating between two reservoirs has efficiency $\frac{1}{3}$ . When the temperature of the cold reservoir is raised by $x$ , its efficiency decreases to $\frac{1}{6}$ . The value of $x$ , if the temperature of the hot reservoir is $99 ° C$ , will be [2023]

16.5 K
33 K
66 K
62 K

1

2

3

4

(4)

$T_{H} = 99 ° C = 99 + 273 = 372 K$

$1 - \frac{T_{H}}{T_{C}} = \frac{1}{3}$

$\frac{T_{C}}{T_{H}} = \frac{2}{3} ...(1)$

$\Rightarrow T_{C} = \frac{2}{3} \times 372 = 2 \times 124 = 248 K$

$1 - \frac{T_{C} + X}{T_{H}} = \frac{1}{6}$

$\Rightarrow \frac{5}{6} = \frac{248 + X}{372}$

$248 + X = 5 \times 62$

$\Rightarrow X = 310 - 248 = 62 K$

Q 6 :

Work done by a Carnot engine operating between temperatures 127°C and 27°C is 2 kJ. The amount of heat transferred to the engine by the reservoir is [2023]

8 kJ
4 kJ
2.67 kJ
2 kJ

1

2

3

4

(1)

$Efficiency of Carnot engine$

$η = 1 - \frac{T_{2}}{T_{1}} = \frac{W}{Q_{1}}$ $\Rightarrow \frac{W}{Q_{1}} = 1 - \frac{300}{400} = \frac{1}{4}$

$\Rightarrow \frac{2 kJ}{Q_{1}} = \frac{1}{4}$

$\Rightarrow Q_{1} = 8 kJ$

Q 7 :

An engine operating between the boiling and freezing points of water will have [2023]

1. efficiency more than 27%

2. efficiency less than the efficiency of a Carnot engine operating between the same two temperatures

3. efficiency equal to 27%

4. efficiency less than 27%

2, 3 and 4 only
2 and 3 only
2 and 4 only
1 and 2 only

1

2

3

4

(3)

$η = (1 - \frac{273}{373}) \times 100 = 26.8 %$

Q 8 :

The volume of an ideal gas increases 8 times and temperature becomes ${(\frac{1}{4})}^{th}$ of the initial temperature during a reversible change. If there is no exchange of heat in this process $(Δ Q = 0)$ , then identify the gas from the following options (assuming gases given in the options are ideal gases): [2026]

${CO}_{2}$
${NH}_{3}$
$O_{2}$
$He$

1

2

3

4

(4)

Topic Question Set

Q 1 :

Water of mass m gram is slowly heated to increase the temperature from $T_{1}$ to $T_{2}$ . The change in entropy of the water, given specific heat of water is $1 {Jkg}^{- 1} K^{- 1}$ , is [2025]

Q 3 :

A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of the sink same, the new temperature of the source will be [2023]

Q 5 :

A Carnot engine operating between two reservoirs has efficiency $\frac{1}{3}$ . When the temperature of the cold reservoir is raised by $x$ , its efficiency decreases to $\frac{1}{6}$ . The value of $x$ , if the temperature of the hot reservoir is $99 ° C$ , will be [2023]

Q 6 :

Work done by a Carnot engine operating between temperatures 127°C and 27°C is 2 kJ. The amount of heat transferred to the engine by the reservoir is [2023]

Q 7 :

An engine operating between the boiling and freezing points of water will have [2023]

1. efficiency more than 27%

2. efficiency less than the efficiency of a Carnot engine operating between the same two temperatures

3. efficiency equal to 27%

4. efficiency less than 27%

Want Us to Email you About Special Offers & Updates?