6. (JEE Main 2023 (Online) 24th January Morning Shift )

1 g of a liquid is converted to vapour at 3 $\times$ 10 $^{5}$ Pa pressure. If 10% of the heat supplied is used for increasing the volume by 1600 cm $^{3}$ during this phase change, then the increase in internal energy in the process will be :

(A) 4800 J

(B) 4320 J

(C) 432000 J

(D) 4.32 $\times$ 10 $^{8}$ J

Correct answer is (B)

Work done = P $Δ$ V

= 3 × 10⁵ × 1600 × 10^–6

= 480 J

Only 10% of heat is used in work done.

Hence $Δ$ Q = 4800 J The rest goes in internal energy, which is 90% of heat.

Change in internal energy = 0.9 × 4800 = 4320 J

7. (JEE Main 2022 (Online) 26th July Morning Shift )

7 mol of a certain monoatomic ideal gas undergoes a temperature increase of $40 K$ at constant pressure. The increase in the internal energy of the gas in this process is :

(Given $R = 8.3 {JK}^{- 1} {mol}^{- 1}$ )

(A) 5810 J

(B) 3486 J

(C) 11620 J

(D) 6972 J

Correct answer is (B)

$Δ U = n C_{v} Δ T$

$= 7 \times \frac{3 R}{2} \times 40$

$= 3486 J$

8. (JEE Main 2022 (Online) 29th June Morning Shift )

A cylinder of fixed capacity of 44.8 litres contains helium gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by 20.0 $^{\circ}$ C will be :

(Given gas constant R = 8.3 JK^{$-$ 1}-mol^{$-$ 1})

(A) 249 J

(B) 415 J

(C) 498 J

(D) 830 J

Correct answer is (C)

$Δ Q = n C_{v} Δ T$ (Isochoric process)

$= 2 \times \frac{3 R}{2} \times 20$

$= 498$ J

9. (JEE Main 2022 (Online) 28th June Evening Shift )

A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the gas during the part BC. The internal energy of the gas at A is 1560 J. The workdone by the gas during the part CA is :

JEE Main 2022 (Online) 28th June Evening Shift Physics - Heat and Thermodynamics Question 105 English

(A) 20 J

(B) 30 J

(C) $-$ 30 J

(D) $-$ 60 J

Correct answer is (C)

$Δ$ U_AB = 40 J as process is isochoric.

$Δ$ U_BC = + 50 (W_BC = $-$ 50 J)

U_C = U_A + $Δ$ U_AB + $Δ$ U_BC = 1650

For CA process,

Q_CA = $-$ 60 J

$Δ$ U_CA + W_CA = $-$ 60

$-$ 90 + W_CA = $-$ 60

$\Rightarrow$ W_CA = +30 J

The graph given is inconsistent with the statement BC may be adiabatic and CA cannot be like isobaric as shown, as increasing volume while rejecting heat at same time.

10. (JEE Main 2022 (Online) 25th July Evening Shift )

A block of ice of mass 120 g at temperature 0 $^{\circ}$ C is put in 300 g of water at 25 $^{\circ}$ C. The x g of ice melts as the temperature of the water reaches 0 $^{\circ}$ C. The value of x is _____________.

[Use specific heat capacity of water = 4200 Jkg^{$-$ 1}K^{$-$ 1}, Latent heat of ice = 3.5 $\times$ 10⁵ Jkg^{$-$ 1}]

Correct answer is (90)

Heat lost by water = Heat gained by ice

$0.3 \times 4200 \times 25 = x \times 3.5 \times 10^{5}$

$x = \frac{0.3 \times 4200 \times 25}{3.5 \times 10^{5}}$

$= 90 \times 100 \times 10^{5} \times 10^{3}$ gram = 90 gm