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QUESTION: 1

**Direction (Q. Nos. 1-4) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.**

**Q. A sample of an ideal gas occupies 20 L under a pressure of 0.5 atm and under isothermal condition. On increasing its pressure to 1.0 atm, its volume will be**

Solution:

Since the system is isothermal, so temperature is constant.

According to Boyle's lawP_{1}V_{1}= P_{2}V_{2}

20×0.5=1×V_{2}

10=V_{2}

QUESTION: 2

A 10.0 cm column of air is trapped by a column of Hg 4.00 cm long in a capillary tube of uniform bore when the tube is held horizontally at 1 atm. Length of the air column when the tube is held vertically with the open end up is

Solution:

V1 = 10xcm³

V2 = ?

P1 = 1atm = 760mmHg

P2 = (760 + 4)mmHg

P1V1 = P2V2

V2 = P1V1 / P2

= 760 × 10x / 764

= 9.5x

Therefore,

= 9.5x / x = 9.5

QUESTION: 3

Volume of an ideal gas is to be decreased by 10% by increase of pressure by x% under isothermal condition. Thus, x is

Solution:

Applying Boyle 's law P1V1=P2V2 as temp is constant.

So let initial pressure =P initial volume =V

Now final volume V2=V-10%of V =9V/10

PV=P2×9V/10 so P2=10P/9

Therefore increment in Pressure=P/9P×100%=100/9

QUESTION: 4

Density of an ideal gas at 298 K and 1.0 atm is found to be 1.25 kg m^{-3}. Density of the gas at 1.5 atm and at 298 K is

Solution:

We know ideal gas equation,

PV=nRT.

n= m(mass)/M(molecular weight).

PV=m/M RT.

PM=m/v RT.

density (d)=m(mass)/v (volume).

PM=dRT.

d=PM/RT.

here, density is directly proportional to pressure and inversely proportional to temperature...

d_{1}/d_{2}=P_{1} T_{2} / P_{2} T_{1}.

1.25/d_{2}= 1×298/1.5×298.

1.25/d_{2}=1/1.5.

d_{2}= 1.25×1.5.

d_{2}=1.875

QUESTION: 5

**Direction (Q. Nos. 5-9) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THANT ONE is correct.**

**Q. An ideal gas is at 300 K and 1 atm. Then volume remains constant if**

Solution:

We know that

pV = nRT

Or V = nRT/p

If we increase both T and P 2 times, only then the volume remains constant.

QUESTION: 6

The pressure of a 1:4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. What would be the partial pressure of dioxygen?

Solution:

Let the number of moles of dihydrogen and dioxygen be 1 and 4.

*Multiple options can be correct

QUESTION: 7

Select correct matching of laws with their corresponding relations.

Solution:

QUESTION: 8

Charle's law defines value of 'k' as

Solution:

According to Charles law,at constant pressure, V∝T

Or V/T = k

*Multiple options can be correct

QUESTION: 9

Consider gases confined by a liquid, as shown in figure.

Density of the liquid = d (all values in SI unit)

**Then **

Solution:

In this setup

The liquid in tube 1 has got lowered by height h1 which means that the pressure inside the tube is greater than the pressure exerted by the atmosphere. This excess pressure must be equal to the pressure exerted by h1 column of liquid of the same cross section ,therefore

Pressure inside tube 1=atmospheric pressure+h1dg

In tube 2 since the level of water is same as that outside the tube so her

Pressure in tube 2=atmospheric pressure

In tube 3 the liquid has risen by height be which means that pressure inside the tube is less than the pressure exerted by atmospheric pressure .

Atmospheric pressure=pressure in tube+h_{3}dg

Pressure in tube=atmospheric pressure -h_{3}dg

QUESTION: 10

**Direction (Q. Nos. 10-11) This section contains a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d)**.

For Charles’ law, graphical representation between log V and log T under isobaric condition is shown below for equation V = kT.

**Q. Relation between p _{1}, p_{2 }and p_{3} is**

Solution:

From ideal gas equation, we have

pV = nRT

Or V = nRT/p

Taing log on both sides

log V = log nRT - log p

Or y = mx - c

Let us make a point constant on time axis like this

Now we have the temperature constant.So to have maximum value of log V, we need to have minimum value of log p. From the graph we can see that we have least pressure for p3 . So the order will be p_{3}<p_{2}<p_{1}.

QUESTION: 11

For Charles’ law, graphical representation between log V and log T under isobaric condition is shown below for equation V = kT

**Q. Volume V = constant k, when**

Solution:

-272deg.

For C = 1 K

V / 1 = K

V = K

QUESTION: 12

**Direction (Q. Nos. 12) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.**

**Q.
G**

Solution:

From Boyle’s law;

p ∝ 1/V ⇒ Graph i (Straight line passing through origin)

Or pV = Constant ⇒ Graph ii (rectangular hyperbola)

Taking log on both side;

log(pV) = logK

logp + logV = k’

logp = -logV = k’ (y = -mx+c) ⇒ Graph iii

log p = log V-1 + k’

Logp = log(1/V) + k’ ⇒ Graph iv

So option c is correct

*Answer can only contain numeric values

QUESTION: 13

**Direction (Q. Nos. 13 and 14) This section contains 2 questions. when worked out will result in an integer from 0 to 9 (both inclusive)**.

**Isothermally compressibility of an ideal gas is **

**Q.
What is the value of a at 0.2 bar?**

Solution:

α = -1/V(dV/dP)_{T}

pV = RT;

V = RT/p

dV/dp = -RT/p^{2}

α = -1/(RT/p)×(-RT/p^{2})

α = 1/p

Or α = (since p = 0.2)

*Answer can only contain numeric values

QUESTION: 14

Cubic expansion coefficient (β) for an ideal gas is

**Q.
Its value at 250 K is x * 10 ^{-3} K ^{-1}. What is value of x?**

Solution:

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