1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

Questions Asked from Heat and Thermodynamics

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