1

JEE Main 2016 (Online) 9th April Morning Slot

The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is :
A
${3 \over 5}$
B
${2 \over 3}$
C
${3 \over 2}$
D
${2 \over 5}$

Explanation

In an isobaric process,

Heat supplied, Q = n Cp $\Delta$ T

Work done, w = nR$\Delta$T

$\therefore$   Ratio = ${w \over Q}$ = ${{nR\Delta T} \over {n{C_p}\Delta T}}$

=   ${R \over {{5 \over 2}R}}$

=   ${2 \over 5}$

[Cp = ${5 \over 2}$ R for monoatomic gas]
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