1

### JEE Main 2019 (Online) 10th January Evening Slot

Two kg of a monoatomic gas is at a pressure of 4 $\times$ 104 N/m2. The density of the gas is 8 kg/m3. What is the order of energy of the gas due to its thermal motion ?
A
104 J
B
103 J
C
105 J
D
106 J

## Explanation

Thermal energy of N molecule

= N$\left( {{3 \over 2}kT} \right)$

= ${N \over {{N_A}}}{3 \over 2}$RT

= ${3 \over 2}$(nRT)

= ${3 \over 2}$PV

= ${3 \over 2}$P$\left( {{m \over 8}} \right)$

= ${3 \over 2}$ $\times$ 4 $\times$ 104 $\times$ ${2 \over 8}$

= 1.5 $\times$ 104

order will 104
2

### JEE Main 2019 (Online) 10th January Evening Slot

An unknown metal of mass 192 g heated to a temperature of 100oC was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4oC. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5oC. (Specific heat of brass is 394 J kg–1 K–1)
A
458 J kg–1 K–1
B
1232 J kg–1 K–1
C
654 J kg–1 K–1
D
916 J kg–1 K–1

## Explanation

192 $\times$ S $\times$ (100 $-$ 21.5)

= 128 $\times$ 394 $\times$ (21.5 $-$ 8.4)

+ 240 $\times$ 4200 $\times$ (21.5 $-$ 8.4)

$\Rightarrow$   S = 916
3

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

A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. considering only translational and rotational modes, the total internal energy of the system is :
A
12 RT
B
20 RT
C
4 RT
D
15 RT

## Explanation

U $= {{{f_1}} \over 2}{n_1}RT + {{{f_2}} \over 2}{n_2}RT$

$= {5 \over 2}\left( {3RT} \right) + {3 \over 2} \times 5RT$

U $= 15RT$
4

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

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TVx = constant, then x is :
A
${5 \over 3}$
B
${2 \over 5}$
C
${3 \over 5}$
D
${2 \over 3}$

## Explanation

For adiabatic process : TV$\gamma $$-1 = constant For diatomic process : \gamma$$-$1 = ${7 \over 5} - 1$

$\therefore$  x = ${2 \over 5}$