1

### JEE Main 2019 (Online) 11th January Evening Slot

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is increased by $\Delta$T. The amount of heat absorbed by gas is (R is gas constant) :
A
${1 \over 2}$ KR$\Delta$T
B
${1 \over 2}$ R$\Delta$T
C
${3 \over 2}$ R$\Delta$T
D
${2K \over 3}$ $\Delta$T

## Explanation

VT = K

$\Rightarrow$  V$\left( {{{PV} \over {nR}}} \right)$ = k $\Rightarrow$ PV2 = K

$\because$  C = ${R \over {1 - x}} +$ Cv (For polytropic process)

C = ${R \over {1 - 2}} + {{3R} \over 2}$ = ${R \over 2}$

$\therefore$  $\Delta$Q = nC $\Delta$T

= ${R \over 2} \times \Delta$T
2

### JEE Main 2019 (Online) 12th January Morning Slot

An ideal gas occupies a volume of 2m3 at a pressure of 3 $\times$ 106 Pa. The energy of the gas is :
A
6 $\times$ 104 J
B
9$\times$ 106 J
C
3 $\times$ 102 J
D
108 J

## Explanation

Energy = ${1 \over 2}$ nRT = ${f \over 2}$PV

= ${f \over 2}$ (3 $\times$ 106) (2)

= f $\times$ 3 $\times$ 106

Considering gas is monoatomic i.e. f = 3

E. = 9 $\times$ 106 J
3

### JEE Main 2019 (Online) 12th January Morning Slot

A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K1 and the of the outer cylinder is K2. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is :
A
K1 + K2
B
${{{K_1} + 3{K_2}} \over 4}$
C
${{{K_1} + {K_2}} \over 2}$
D
${{2{K_1} + 3{K_2}} \over 5}$

## Explanation Keq = ${{{K_1}{A_1} + {K_2}{A_2}} \over {{A_1} + {A_2}}}$

= ${{{K_1}\left( {\pi {R^2}} \right) + {K_2}\left( {3\pi {R^2}} \right)} \over {4\pi {R^2}}}$

= ${{{K_1} + 3{K_2}} \over 4}$
4

### JEE Main 2019 (Online) 12th January Evening Slot

An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature 300 K. The mean time between two successive collisions is 6 $\times$ 10–8 s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to
A
0.5 $\times$ 10$-$8 s
B
4 $\times$ 10$-$8 s
C
3 $\times$ 10$-$6 s
D
2 $\times$ 10$-$7 s

## Explanation

t $\propto$ ${{Volume} \over {velocity}}$

volume $\propto$ ${T \over P}$

$\therefore$  t $\propto$ ${{\sqrt T } \over P}$

${{{t_1}} \over {6 \times {{10}^{ - 8}}}} = {{\sqrt {500} } \over {2P}} \times {P \over {\sqrt {300} }}$

t1 = 3.8 $\times$ 10$-$8

$\approx$ 4 $\times$ 10$-$8