# NEET Questions Solved

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NEET - 2015

Figure below shows two paths that may be taken by a gas to go from a state A to a state C. In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be-

(a) 380 J
(b) 500 J
(c) 460 J
(d) 300 J

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NEET - 2012

A thermodynamic system is taken through the cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is

(a)2 pV                                     (b)4 pV

(c)$\frac{1}{2}pV$                                    (d)pV

For given cyclic process, $∆U=0$

Also, W= -area enclosed by the curve

=$AB×AD$

=$\left(2p-p\right)\left(3V-V\right)$

=$-p×2V$

Difficulty Level:

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NEET - 2012

An ideal gas goes from state A to state B

via three different processes as indicated

in the p-V diagram

If  indicates the heat absorbed

by the gas along the three processes and

indicates the change in

internal energy along the three processes

respectively, then

(a)

(b)

(c)

(d)

For all process 1, 2 and 3

$∆\mathrm{U}={\mathrm{U}}_{\mathrm{B}}-{\mathrm{U}}_{\mathrm{A}}$

so,      $∆{\mathrm{U}}_{1}=∆{\mathrm{U}}_{2}=∆{\mathrm{U}}_{3}$

Now,   $∆\mathrm{Q}=∆\mathrm{U}+∆\mathrm{W}$

Now, $∆\mathrm{W}$= work done by the gas

so, $∆{\mathrm{Q}}_{1}>∆{\mathrm{Q}}_{2}>∆{\mathrm{Q}}_{3}$

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NEET - 2009

The internal energy change in a system that has absorbed 2 kcal of heat and done 500 J of work is

(a) 8900 J                                      (b) 6400 J

(c) 5400 J                                      (d) 7900 J

Heat given to a system ($∆Q$) is equal to the sum of increase in the internal energy $\left(∆u\right)$ and the work done $\left(∆W\right)$ by the system against the surrounding and 1 cal=4.2J.

According to first law of thermodynamics

$∆U=Q-W$

= $2×4.2×1000-500$

= $8400-500$

= 7900 J

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A system performs work ΔW when an amount of heat is ΔQ added to the system, the corresponding change in the internal energy is ΔU. A unique function of the initial and final states (irrespective of the mode of change) is -

(1) ΔQ

(2) ΔW

(3) ΔU and ΔQ

(4) ΔU

(4) Change in internal energy (ΔU) depends upon initial an find state of the function while ΔQ and ΔW are path dependent also.

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A container of volume 1m3 is divided into two equal compartments by a partition. One of these compartments contains an ideal gas at 300 K. The other compartment is vaccum. The whole system is thermally isolated from its surroundings. The partition is removed and the gas expands to occupy the whole volume of the container. Its temperature now would be -

(1) 300 K

(2) 239 K

(3) 200 K

(4) 100 K

(1) This is the case of free expansion and in this case $\Delta W=0$, $\Delta U=0$ so temperature remains same i.e. 300 K.

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110 J of heat is added to a gaseous system, whose internal energy change is 40 J, then the amount of external work done is

(1) 150 J

(2) 70 J

(3) 110 J

(4) 40 J

(2) $\Delta Q=\Delta U+\Delta W$

$\Delta W=\Delta Q-\Delta U=100-40=70J$

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When the amount of work done is 333 cal and change in internal energy is 167 cal, then the heat supplied is -

(1) 166 cal

(2) 333 cal

(3) 500 cal

(4) 400 cal

(3) $\Delta Q=\Delta U+\Delta W=167+333=500\text{\hspace{0.17em}}cal$

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