1

### 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)
2

### 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 3 MCQ (Single Correct Answer) ### 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 4 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 15th April Morning Slot A Carnot's engine works as a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is : A 420 J B 772 J C 2100 J D 2520 J ## Explanation Given, Cold body temperature (T2) = 250 K Hot body temperature (T1) = 300 K Received heat (Q2) = 500 cal Let, required works done = \omega \therefore\,\,\,\, For the refrigerator, Efficiency = 1 - {{T2} \over {T1}} = {\omega \over {{Q_2} + \omega }} \Rightarrow$$\,\,\,\,$ 1 $-$ ${{250} \over {300}} = {\omega \over {{Q_2} + \omega }}$

$\Rightarrow $$\,\,\,\, {{{Q_2} + \omega } \over \omega } = {{300} \over {50}} = 6 \Rightarrow$$\,\,\,\,$ $\omega = {{{Q_2}} \over 5}$

$\Rightarrow $$\,\,\,\, \omega = {{500 \times 4.2} \over 5} \Rightarrow$$\,\,\,\,$ $\omega = 420\,J$