1

### 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

### 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

### 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

### 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