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1. ⇒ (NEET 2018 )

A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 10⁵ N m^–2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is

(A) 104.3 J

(B) 208.7 J

(C) 42.2 J

(D) 84.5 J

Correct answer is (B)

Using first law of thermodynamics,

$Δ$ Q = $Δ$ U + $Δ$ W

$\Rightarrow$ 54 × 4.18 = $Δ$ U + 1.013 × 10⁵(167.1 × 10^–6 – 0)

$\Rightarrow$ $Δ$ U = 208.7 J

2. ⇒ (AIPMT 2015 Cancelled Paper )

Figure below shows two paths that may be taken by a gas to go from a state A to a state C.

AIPMT 2015 Cancelled Paper Physics - Heat and Thermodynamics Question 59 English
In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be

(A) 460 J

(B) 300 J

(C) 380 J

(D) 500 J

Correct answer is (A)

AIPMT 2015 Cancelled Paper Physics - Heat and Thermodynamics Question 59 English Explanation
Considering the cyclic process ABCA
Q_cyclic = W = area of $Δ$ ABC

$Q_{A B} + Q_{B C} + Q_{C A} = \frac{1}{2} \times B C \times A B$

$400 + 100 + Q_{C A} = \frac{1}{2} \times (2 \times 10^{- 3}) \times 4 \times 10^{4}$

$400 + 100 - Q_{A C} = 40$

Hence, Q_AC = 460 J

3. ⇒ (AIPMT 2014 )

A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by the system in the cycle is

AIPMT 2014 Physics - Heat and Thermodynamics Question 57 English

(A) P₀V₀

(B) 2P₀V₀

(C) $\frac{P_{0} V_{0}}{2}$

(D) zero

Correct answer is (D)

AIPMT 2014 Physics - Heat and Thermodynamics Question 57 English Explanation In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise. As is clear from figure,

$W_{A E D A} = + a r e a o f Δ A E D = + \frac{1}{2} P_{0} V_{0}$

$W_{B C E B} = - a r e a o f Δ B C E = - \frac{1}{2} P_{0} V_{0}$

The net work done by the system is

$W_{n e t} = W_{A E D A} + W_{B C E B}$

$= + \frac{1}{2} P_{0} V_{0} - \frac{1}{2} P_{0} V_{0} = 0$

4. ⇒ (NEET 2013 (Karnataka) )

A system is taken from state a to state c by two paths adc and abc as shown in the figure. The internal energy at a is $U_{a} = 10 J$ . Along the path adc the amount of heat absorbed dQ₁ $=$ 50 J and the work obtained dW₁ $=$ 20 J whereas along the path abc the heat absorbed dQ₂ = 36 J. The amount of work allong the path abc is

NEET 2013 (Karnataka) Physics - Heat and Thermodynamics Question 51 English

(A) 10 J

(B) 12 J

(C) 36 J

(D) 6 J

Correct answer is (D)

From first law of thermodynamics

$Q_{a d c} = Δ U_{a d c} + W_{a d c}$

$50 J = Δ U_{a d c} + 20 J$

$Δ U_{a d c} = 30 J$

Again

$Q_{a b c} = Δ U_{a b c} + W_{a b c}$

$W_{a b c} = Q_{a b c} - Δ U_{a b c}$

$W_{a b c} = 36 J - 30 J$

$W_{a b c} = 6 J$

5. ⇒ (NEET 2013 )

The amount of heat energy required to raise the temperature of 1 g of Helium at NTP, from T₁K to T₂K is

(A) $\frac{3}{4} N_{a} k_{B} (T_{2} - T_{1})$

(B) $\frac{3}{4} N_{a} k_{B} (\frac{T_{2}}{T_{1}})$

(C) $\frac{3}{8} N_{a} k_{B} (T_{2} - T_{1})$

(D) $\frac{3}{2} N_{a} k_{B} (T_{2} - T_{1})$

Correct answer is (C)

From first law of thermodynamics

$Δ Q = Δ U + Δ W = \frac{3}{2} . \frac{1}{4} R (T_{2} - T_{1}) + 0$

$= \frac{3}{8} N_{a} K_{B} (T_{2} - T_{1})$ $[∵ K = \frac{R}{N}]$

6. ⇒ (NEET 2013 )

A gas is taken through the cycle
A $\to$ B $\to$ C $\to$ A, as shown. what is the net work done by the gas?

NEET 2013 Physics - Heat and Thermodynamics Question 54 English

(A) Zero

(B) $-$ 2000 J

(D) 1000 J

Correct answer is (D)

W_net = Area of triangle ABC

$= \frac{1}{2} A C \times B C$

$= \frac{1}{2} \times 5 \times 10^{- 3} \times 4 \times 10^{5} = 1000 J$

7. ⇒ (AIPMT 2012 Mains )

An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram.

AIPMT 2012 Mains Physics - Heat and Thermodynamics Question 46 English

If Q₁, Q₂, Q₃ indicate the heat absorbed by the gas along the three processes and $Δ$ U₁, $Δ$ U₂, $Δ$ U₃ indicate the change in internal energy along the three processes respectively, then

(A) Q₁ > Q₂ > Q₃ and $Δ$ U₁ = $Δ$ U₂ = $Δ$ U₃

(B) Q₃ > Q₂ > Q₁ and $Δ$ U₁ = $Δ$ U₂ = $Δ$ U₃

(C) Q₁ = Q₂ = Q₃ and $Δ$ U₁ > $Δ$ U₂ > $Δ$ U₃

(D) Q₃ = Q₂ = Q₁ and $Δ$ U₁ > $Δ$ U₂ > $Δ$ U₃

Correct answer is (A)

Initial and final condition is same for all process

$Δ U_{1} = Δ U_{2} = Δ U_{3}$

from first law of thermodynamics

$Δ Q = Δ U + Δ W$

Work done

$Δ W_{1} > Δ W_{2} > Δ W_{3}$ (Area of P.V. graph)

So $Δ Q_{1} > Δ Q_{2} > Δ Q_{3}$

8. ⇒ (AIPMT 2012 Prelims )

A thermodynamic system is taken through the cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is

AIPMT 2012 Prelims Physics - Heat and Thermodynamics Question 23 English

(A) 2PV

(B) 4PV

(C) $\frac{1}{2}$ PV

(D) PV

Correct answer is (A)

AIPMT 2012 Prelims Physics - Heat and Thermodynamics Question 23 English Explanation
In a cyclic process,

$Δ U = 0$

In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise.

$∴ Δ$ W = – Area of rectangle ABCD = – P(2V) = – 2PV

According to first law of thermodynamics

$Δ Q = Δ U + Δ W \Rightarrow Δ Q = Δ W$
(As $Δ U = 0$ )

i.e., heat supplied to the system is equal to the work done

So heat absorbed, $Δ$ Q = $Δ$ W = – 2PV

$∴$ Heat rejected by the gas = 2PV

9. ⇒ (AIPMT 2009 )

The internal energy change in a system that has absorbed 2 kcal of heat and done 500 J of work is

(A) 6400 J

(B) 5400 J

(C) 7900 J

(D) 8900 J

Correct answer is (C)

When a quantity Q of heat is supplied to a system it is used to do an amount of work W by the system and to increase the internal energy of the system by ∆U :

Q = ∆U + W

Here, Q given in kilo calories is converted into joule.
Therefore Q = 2×1000×4.2 J = 8400 J

The increase the internal energy

∆U = Q – W = 8400 – 500 = 7900 J

⇒Thermodynamics

Topic 01: First Law of Thermodynamic