1

JEE Main 2017 (Online) 9th April Morning Slot

N moles of a diatomic gas in a cylinder are at a temperature T. Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get converted into monoatomic gas. What is the change in the total kinetic energy of the gas ?
A
${1 \over 2}$ nRT
B
0
C
${3 \over 2}$ nRT
D
${5 \over 2}$ nRT

Explanation

Initial kinetic energy of N mole of diatomic gas,

Ki = N${5 \over 2}$ RT

Kinetic energy of n mole of monoatomic gas
= n ${3 \over 2}$ RT

When n mole of diatomic gas converted into monoatomic gas then remaining diatomic gas = (N $-$ n)

Find kinetic energy,

KF = (2m)${3 \over 2}$RT + (N $-$ n)${5 \over 2}$ RT

= ${1 \over 2}$ nRT + ${5 \over 2}$ NRT

$\therefore\,\,\,$ Change in kinetic energy,

$\Delta$K = Kf $-$ Ki = ${1 \over 2}$ nRT
2

JEE Main 2017 (Online) 9th April Morning Slot

For the P-V diagram given for an ideal gas,

out of the following which one correctly represents the T-P diagram ?
A
B
C
D

Explanation

We know,

PV = nRT

here R is a constant,

assuming n = number of moles does not change, we get,

PV = KT, K = constant.

Given that, PV = constant,

So, T = constant, Hence the process is isothermol.

From the graph you can see,

Pressure at point 1 is higher than that at point 2. So correct answer is (c)
3

JEE Main 2018 (Offline)

Two moles of an ideal monatomic gas occupies a volume V at 27oC. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.
A
(a) 195 K (b) 2.7 kJ
B
(a) 189 K (b) 2.7 kJ
C
(a) 195 K (b) –2.7 kJ
D
(a) 189 K (b) – 2.7 kJ

Explanation

pv$\gamma$ = constant.

and we know, pv = nRT

$\therefore\,\,\,$ p = ${{nRT} \over v}$

$\therefore\,\,\,$ ${{nRT} \over v} \times {v^\gamma }$ = constant

$\Rightarrow $$\,\,\, T v\gamma$$-$1 = constant.

$\therefore\,\,\,$ T1 v1$\gamma $$-1 = T2 v2\gamma$$-$1

Given that,

This is a monoatomic gas.

So, degree of frequency f = 3

$\therefore\,\,\,$ $\gamma$ = ${{{c_p}} \over {{c_v}}}$ = 1 + ${2 \over f}$ = 1 + ${2 \over 3}$ = ${5 \over 3}$

T1 = 27 + 273 = 300 K

v1 = v

and v2 = 2v

$\therefore\,\,\,$ T2 (2V)$^{{2 \over 3}}$ = 300(v)$^{{2 \over 3}}$

$\Rightarrow$$\,\,\,$ T2 = ${{300} \over {{2^{{2 \over 3}}}}}$ = 189 K

Change in internal energy,

$\Delta$U = ${1 \over 2}$ nfR$\Delta$T

= ${1 \over 2}$ $\times$ 2 $\times$ 3 $\times$ 8.31 $\times$ (189 $-$ 300)

= $-$ 2.7 KJ
4

JEE Main 2018 (Offline)

The mass of a hydrogen molecule is 3.32 $\times$ 10-27 kg. If 1023 hydrogen molecules strike, per second, a fixed wall of area 2 cm2 at an angle of 45o to the normal, and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly:
A
2.35 $\times$ 103 N m-2
B
4.70 $\times$ 103 N m-2
C
2.35 $\times$ 102 N m-2
D
4.70 $\times$ 102 N m-2

Explanation

Considering one hydrogen molecule :

As collision is elastic so, e = 1

Initial momentum,

$\overrightarrow {{P_i}}$ = ${{mv} \over {\sqrt 2 }}$ $\widehat i$ $-$ ${{mv} \over {\sqrt 2 }}$ $\widehat j$

Final momentum,

$\overrightarrow {{P_f}}$ = ${{mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)$ $-$ ${{mv} \over {\sqrt 2 }}\widehat j$

$\therefore\,\,\,$ Change in momentum for single H molecule,

$\Delta$P = $\overrightarrow {{P_f}}$ $-$ $\overrightarrow {{P_i}}$

= ${{2mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)$

$\therefore\,\,\,$ $\left| {\Delta P} \right|$ = ${{2mv} \over {\sqrt 2 }}$

Now for n hydrogen molecule total momentum changes per second,

= $\left( {{{2mv} \over {\sqrt 2 }}} \right)$ $\times$ n

As we know,

Force (F) = ${{\Delta P} \over {\Delta t}}$

= ${{{2mv} \over {\sqrt 2 }}}$ $\times$ n

$\therefore\,\,\,$ As direction of $\Delta$P is towards negative i so force on the molecule will also be towards negative i direction. From Newton's third law, the reaction force will be on the wall in positive i direction with same magnitude.

$\therefore\,\,\,$ Force on the wall = ${{{2mv} \over {\sqrt 2 }}}$ $\times$ n

$\therefore\,\,\,$ Pressure on the wall, P = ${F \over A}$

= ${{2mv\,n} \over {\sqrt 2 A}}$

= ${{2 \times 3.32 \times {{10}^{ - 27}} \times {{10}^3} \times {{10}^{23}}} \over {\sqrt 2 \times2\times {{10}^{ - 4}}}}$

= 2.35 $\times$ 103 N m$-$2