1
MCQ (Single Correct Answer)

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)
2
MCQ (Single Correct Answer)

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

For adiabatic process,

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
3
MCQ (Single Correct Answer)

JEE Main 2018 (Offline)

The mass of a hydrogen molecule is 3.32 $$\times$$ 10-27 kg. If 1023 hydrogen molecules strike, per second, a fixed wall of area 2 cm2 at an angle of 45o to the normal, and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly:
A
2.35 $$\times$$ 103 N m-2
B
4.70 $$\times$$ 103 N m-2
C
2.35 $$\times$$ 102 N m-2
D
4.70 $$\times$$ 102 N m-2

Explanation

Considering one hydrogen molecule :



As collision is elastic so, e = 1

Initial momentum,

$$\overrightarrow {{P_i}} $$ = $${{mv} \over {\sqrt 2 }}$$ $$\widehat i$$ $$-$$ $${{mv} \over {\sqrt 2 }}$$ $$\widehat j$$

Final momentum,

$$\overrightarrow {{P_f}} $$ = $${{mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)$$ $$-$$ $${{mv} \over {\sqrt 2 }}\widehat j$$

$$\therefore\,\,\,$$ Change in momentum for single H molecule,

$$\Delta $$P = $$\overrightarrow {{P_f}} $$ $$-$$ $$\overrightarrow {{P_i}} $$

= $${{2mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)$$

$$\therefore\,\,\,$$ $$\left| {\Delta P} \right|$$ = $${{2mv} \over {\sqrt 2 }}$$

Now for n hydrogen molecule total momentum changes per second,

= $$\left( {{{2mv} \over {\sqrt 2 }}} \right)$$ $$ \times $$ n

As we know,

Force (F) = $${{\Delta P} \over {\Delta t}}$$

= $${{{2mv} \over {\sqrt 2 }}}$$ $$ \times $$ n

$$\therefore\,\,\,$$ As direction of $$\Delta $$P is towards negative i so force on the molecule will also be towards negative i direction. From Newton's third law, the reaction force will be on the wall in positive i direction with same magnitude.

$$\therefore\,\,\,$$ Force on the wall = $${{{2mv} \over {\sqrt 2 }}}$$ $$ \times $$ n

$$\therefore\,\,\,$$ Pressure on the wall, P = $${F \over A}$$

= $${{2mv\,n} \over {\sqrt 2 A}}$$

= $${{2 \times 3.32 \times {{10}^{ - 27}} \times {{10}^3} \times {{10}^{23}}} \over {\sqrt 2 \times2\times {{10}^{ - 4}}}}$$

= 2.35 $$ \times $$ 103 N m$$-$$2
4
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Morning Slot

A Carnot's engine works as a refrigerator between $$250$$ K and $$300$$ K. It receives $$500$$ cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is :
A
$$420$$ $$J$$
B
$$772$$ $$J$$
C
$$2100$$ $$J$$
D
$$2520$$ $$J$$

Explanation

Given,

Cold body temperature (T2) = 250 K

Hot body temperature (T1) = 300 K

Received heat (Q2) = 500 cal

Let, required works done = $$\omega $$

$$\therefore\,\,\,\,$$ For the refrigerator,

Efficiency = $$1 - {{T2} \over {T1}} = {\omega \over {{Q_2} + \omega }}$$

$$ \Rightarrow $$$$\,\,\,\,$$ 1 $$-$$ $${{250} \over {300}} = {\omega \over {{Q_2} + \omega }}$$

$$ \Rightarrow $$$$\,\,\,\,$$ $${{{Q_2} + \omega } \over \omega }$$ = $${{300} \over {50}}$$ = 6

$$ \Rightarrow $$$$\,\,\,\,$$ $$\omega = {{{Q_2}} \over 5}$$

$$ \Rightarrow $$$$\,\,\,\,$$ $$\omega = {{500 \times 4.2} \over 5}$$

$$ \Rightarrow $$$$\,\,\,\,$$ $$\omega = 420\,J$$

Questions Asked from Heat and Thermodynamics

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