1

### JEE Main 2018 (Online) 16th April Morning Slot

One mole of an ideal monoatomic gas is taken along the path ABCA as show in the PV diagram. The maximum temperature attained by the gas along the path BC is given by : A
${{25} \over {16}}\,{{{P_o}{V_o}} \over R}$
B
${{25} \over {8}}\,{{{P_o}{V_o}} \over R}$
C
${{25} \over {4}}\,{{{P_o}{V_o}} \over R}$
D
${{5} \over {8}}\,{{{P_o}{V_o}} \over R}$

## Explanation

Equation of line BC,

P = P0 $-$ ${{2{P_0}} \over {{V_0}}}\left( {V - 2{V_0}} \right)$

As $\,\,\,\,\,$ PV = nRT

$\therefore\,\,\,\,$ T = ${{PV} \over {nR}}$

= ${{{P_0}V - {{2{P_0}{V_2}} \over {{V_0}}} + 4{P_0}V} \over {1 \times R}}$

(As given n = 1 mole gas)

$\Rightarrow $$\,\,\, T = {{{P_0}} \over R}\left[ {5V - {{2{V^2}} \over {{V_0}}}} \right] For maximum value of T {{dT} \over {dV}} = 0 \therefore\,\,\, 5 - {{4V} \over {{V_0}}} = 0 \Rightarrow$$\,\,\,$ V = ${5 \over 4}{V_0}$

$\therefore\,\,\,$ Tmax = ${{{P_0}} \over R}\left[ {5 \times {{5{V_0}} \over 4} - {2 \over {{V_0}}} \times {{25} \over {16}}{V_0}^2} \right]$

= ${{25} \over 8} \times {{{P_0}\,{V_0}} \over R}$
2

### JEE Main 2019 (Online) 9th January Morning Slot

A mixture of 2 moles of helium gas (atomic mass = 4 u), and 1 mole of argon gas (atomic mass = 40 u) is kept at 300 K in a container. The ratio of their rms speeds $\left[ {{{{V_{rms}}\,(helium)} \over {{V_{rms}}\,(\arg on)}}} \right],$ is close to :
A
3.16
B
0.32
C
0.45
D
2.24

$\mu$

## Explanation

We know,

Vrms = $\sqrt {{{3RT} \over M}}$

Where M = molar mass of the gas.

Here temperature is 300 K for both the gas. So temperature is constant. R is also a constant.

$\therefore$   Vrms $\propto \,\,\sqrt {{1 \over M}}$

$\therefore$    ${{{V_{rms}}(helium|)} \over {{V_{rms}}\left( {\arg an)} \right)}} = \sqrt {{{{M_{Ar}}} \over {{M_{He}}}}}$

= $\sqrt {{{40} \over 4}}$

= $\sqrt {10}$

= 3.16
3

### JEE Main 2019 (Online) 9th January Morning Slot

Temperature difference of 120oC is maintained between ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as AB and length ${{3L} \over 2},$ is connected across AB (see figure). In steady state, temperature difference between P and Q will be close to : A
45oC
B
75oC
C
60oC
D
35oC

## Explanation

We know,

Resistance, R = ${{\rho L} \over A}$

$\therefore$   R $\propto$ L

So, Resistance is directly proportional to lengt5h of the rod. Equivalent resistance between A and B is,

Req = ${R \over 2} + {{R \times {{3R} \over 2}} \over {R + {{3R} \over 2}}} + {R \over 2}$

= R + ${{3{R^2}} \over 2} \times {2 \over {5R}}$

= R + ${{3R} \over 5}$

= ${{8R} \over 5}$

Thermal current between point A and B is,

I = ${{\Delta {T_{AB}}} \over {{{\mathop{\rm R}\nolimits} _{eq}}}}$

= ${{120 - 0} \over {{{8R} \over 5}}}$

= ${{120 \times 5} \over {8R}}$

Resistance between P and Q = ${{3R} \over 5}$

$\therefore$   $\Delta$TPQ = ${\rm I} \times {{3R} \over 5}$

= ${{120 \times 5} \over {8R}} \times {{3R} \over 5}$

= 45oC
4

### JEE Main 2019 (Online) 9th January Morning Slot

A gas can be taken from A to B via two different processes ACB and ADB. When path ACB is used 60 J of heat flows into the system and 30 J of work is done by the system. If path ADB is used work done by the system is 10 J. The heat Flow into the system in path ADB is :
A
40 J
B
80 J
C
100 J
D
20 J

## Explanation

Using law of thermodynamics in path ACB,

$\Delta$QACB = $\Delta$UACB + $\Delta$WACB

$\Rightarrow$   60 = $\Delta$UACB + 30

$\Rightarrow$   $\Delta$UACB = 30 J

As value of $\Delta$U is path independent,

$\therefore$   $\Delta$UACB = $\Delta$UADB = 30 J

$\therefore$   In path ADB,

$\Delta$QADB = $\Delta$UADB + $\Delta$WADB

$\Rightarrow$$\Delta$QADB = 30 + 10 = 40 J