1

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

An engine operates by taking n moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is : (Take Cv = 1.5 R, where R is gas constant) A
0.24
B
0.15
C
0.32
D
0.08

## Explanation

Work done by engine = area under closed curve = P0 V0

According to the principle of calorimetry,

Heat given to the system,

Q = QAB + QBC

= nCv $\Delta$TAB + nCp $\Delta$TBC

= ${3 \over 2}$ (nR TB $-$ nRTA) + ${5 \over 2}$ (nRTc + nRTB)

= ${3 \over 2}$ (2P0 V0 $-$ P0 V0) + ${5 \over 2}$ (4P0 V0 $-$ 2P0 V0)

= ${13 \over 2}$ P0 V0

$\therefore\,\,\,$ Thermal efficiency

= ${W \over Q}$ = ${{{P_0}\,{V_0}} \over {{{13} \over 2}{P_0}{V_0}}}$ = ${2 \over {13}}$ = 0.15
2

### 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

$\therefore\,\,\,$ ${{{C_p}} \over {{C_v}}}$ = ${{{7 \over 2}R} \over {{5 \over 2}R}}$ = ${7 \over 5}$
3

### 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
4

### 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)