1
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  $$ \propto $$ $$\rho $$
2
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
3
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

CP and Cv are specific heats at constant pressure and constant volume respectively. It is observed that
CP – Cv = a for hydrogen gas
CP – Cv = b for nitrogen gas
The correct relation between a and b is
A
a = 28 b
B
a = 1/14 b
C
a = b
D
a = 14 b

Explanation

As we know, for 1 g mole of a gas,

Cp – Cv = R where Cp and Cv are molar specific heat capacities.

So, when n gram moles are given,

Cp – Cv = $${R \over n}$$

For hydrogen (n = 2), Cp – Cv = $${R \over 2}$$ = $$a$$

For nitrogen (n = 28), Cp – Cv = $${R \over 28}$$ = $$b$$

$$ \therefore $$ $${a \over b} = 14$$

$$ \Rightarrow $$ $$a$$ = 14 $$b$$
4
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 8th April Morning Slot

An engine operates by taking n moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is : (Take Cv = 1.5 R, where R is gas constant)

A
0.24
B
0.15
C
0.32
D
0.08

Explanation

Work done by engine = area under closed curve = P0 V0

According to the principle of calorimetry,

Heat given to the system,

Q = QAB + QBC

= nCv $$\Delta $$TAB + nCp $$\Delta $$TBC

= $${3 \over 2}$$ (nR TB $$-$$ nRTA) + $${5 \over 2}$$ (nRTc + nRTB)

= $${3 \over 2}$$ (2P0 V0 $$-$$ P0 V0) + $${5 \over 2}$$ (4P0 V0 $$-$$ 2P0 V0)

= $${13 \over 2}$$ P0 V0

$$\therefore\,\,\,$$ Thermal efficiency

= $${W \over Q}$$ = $${{{P_0}\,{V_0}} \over {{{13} \over 2}{P_0}{V_0}}}$$ = $${2 \over {13}}$$ = 0.15

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