1

### JEE Main 2017 (Online) 8th April Morning Slot

An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure (Cp ) and at constant volume (Cv) is :
A
6
B
${7 \over 2}$
C
${5 \over 2}$
D
${7 \over 5}$

## Explanation

For ideal gas molecule with 5 degree of freedom,

Cv = ${5 \over 2}$ R and Cp = ${7 \over 2}$ R

$\therefore\,\,\,$ ${{{C_p}} \over {{C_v}}}$ = ${{{7 \over 2}R} \over {{5 \over 2}R}}$ = ${7 \over 5}$
2

### 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
3

### 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)
4

### 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