1

### JEE Main 2016 (Online) 9th April Morning Slot

200 g water is heated from 40oC to 60oC. Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water = 4184 J/kg/K) :
A
8.4 kJ
B
4.2 kJ
C
16.7 kJ
D
167.4 kJ

## Explanation

According to the first law of thermodynamics,

Q = $\Delta$u + w

For isochoric process Q = $\Delta$U = ms$\Delta$t

$\Delta$T = (233 $-$ 213) = 20 k

$\therefore$   $\Delta$u = 0.2 $\times$ 4184 $\times$ 20 = 16.7 kJ
2

### JEE Main 2016 (Online) 10th April Morning Slot

A Carnot freezer takes heat from water at 0oC inside it and rejects it to the room at a temperature of 27oC. The latent heat of ice is 336×103 J kg−1. If 5 kg of water at 0oC is converted into ice at 0oC by the freezer, then the energy consumed by the freezer is close to :
A
1.67 $\times$ 105 J
B
1.68 $\times$ 106 J
C
1.51 $\times$ 105 J
D
1.71 $\times$ 107 J

## Explanation Total heat required to freeze 5 kg water,

=   5 $\times$ 336  $\times$ 103 J

=   1680  $\times$  103   Joule

=   1680 kJ

For carnot cycle,

${{{Q_2}} \over {{Q_1}}}$ = ${{{T_2}} \over {{T_1}}}$

$\Rightarrow$   ${{{Q_2}} \over {1680}}$ = ${{300} \over {273}}$

$\Rightarrow$   Q2 = 1680 $\times$  ${{300} \over {273}}$ kJ

$\therefore$   W = Q2 $-$ Q1

=   1680$\left( {{{300} \over {273}} - 1} \right)$

=   1680 $\times$ ${{27} \over {273}}$

=   166.15 kJ

=   166.15 $\times$ 103 J

=   1.66 $\times$ 105 J
3

### JEE Main 2016 (Online) 10th April Morning Slot

Which of the following shows the correct relationship between the pressure ‘P’ and density $\rho$ of an ideal gas at constant temperature ?
A B C D ## Explanation

We know, ideal gas equation,

PV = nRT

Here T = constant.

$\therefore$   PV = constant

$\Rightarrow$   P ${m \over \rho }$ = constant

$\Rightarrow$   P  $\propto$ $\rho$
4

### JEE Main 2017 (Offline)

The temperature of an open room of volume 30 m3 increases from 17oC to 27oC due to the sunshine. The atmospheric pressure in the room remains 1 $\times$ 105 Pa. If Ni and Nf are the number of molecules in the room before and after heating, then Nf – Ni will be :
A
- 1.61 $\times$ 1023
B
1.38 $\times$ 1023
C
2.5 $\times$ 1025
D
- 2.5 $\times$ 1025

## Explanation

Given: Initial temperature Ti = 17 + 273 = 290 K

Final temperature Tf = 27 + 273 = 300 K

Atmospheric pressure, P0 = 1 × 105 Pa

Volume of room, V0 = 30 m3

Difference in number of molecules, Nf – Ni = ?

We know PV = nRT = ${N \over {{N_A}}}$RT

The number of molecules

N = ${{PV{N_A}} \over {RT}}$

$\therefore$ Nf – Ni = ${{{P_0}{V_0}{N_A}} \over R}\left( {{1 \over {{T_f}}} - {1 \over {{T_i}}}} \right)$

= ${{1 \times {{10}^5} \times 30 \times 6.023 \times {{10}^{23}}} \over {8.314}}\left( {{1 \over {300}} - {1 \over {290}}} \right)$

= – 2.5 $\times$ 1025