# 8 - Thermodynamics Questions Answers

A gas expands against a variable external pressure given by **p = 10/V atm**, where **V **is the volume at each stage of expansion. in expanding from **10 L **to **100 L**, the gas undergoes a change in internal energy **Δ****U = 418 J**. How much heat has been absorbed?

**Asked By: RICHIK BANDYOPADHYAY**

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**Joshi sir comment**

process is isothermal because pV = 10 = nRT

so work done W = 2.303nRTlog(V_{2}/V_{1})

so Q = **Δ****U **+ W

A CARNOT ENGINE WORKING B/W 27^{0} C & --123^{0}C HAS EFFICIENCY η . IF TEMP OF SINK IS DECREASED BY 50K THE EFFICIENCY BECOMES??

ANS 4η/3

**Asked By: SARIKA**

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**Joshi sir comment**

according to the given condition η = 1-(150/300) = 0.5

now after decreasing temperature of sink by 50K

we get the new efficiency = 1-(100/300) = 2/3 = 4*0.5/3 = 4η/3

in a thermodynamic process on 2 moles of a monoatomic gas work done on gas is 100J & change in temp is 30 C . The change in internal energy of the gas in this process is??

**Asked By: SARIKA**

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**Joshi sir comment**

change in internal energy = nC_{v}dT = 2*[R/(γ-1)]*30 = 60*8.31/[(5/3)-1]

IN THE PREVIOUS QUES P^{2}V^{3 }=K BUT SORRY I AM NOT GETTING HOW PV^{3/2 }=K

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

take square root on both the sides

DURING AN ADIABATIC PROCESS IF THE PRESSURE OF AN IDEAL GAS IS PROPTIONAL TO THE CUBE OF ITS ABSOLUTE TEMP. THEN RATIO OF SPECIFIC HEATS AT CONSTANT PRESSURE & AT CONST. VOL IS?

ANS 3/2

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

P α T^{3}

so P α (PV)^{3}

so P^{2}V^{3} = K

or PV^{3/2 }= K

so on the basis of PVγ = K calculate γ

Q IF Al DIPPED IN H2O AT 10*C THEN FORCE OF BUOYANCY WILL INCREASE OR DECREASE?

Q TEMP. OF A SOLID OBJECT IS OBSV. CONST .DURING A PERIOD . IN THIS PERIOD

a)HEAT MAY HAVE BEEN SUPPLIED TO IT

b) H MAY HAVE BEEN EXTRACTED FROM IT

c) no H SUPPLIED TO IT

d)NO H EXTRACTED FROM IT

more than one opt is correct . pls explain your each opt with expln.

**Asked By: SARIKA**

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**Joshi sir comment**

It depends that either Al or water will change its density more rapidly, because Bouyancy is based on densities of solid and liquid

Answer : a, because it may be an isothermal process

b, because in isothermal process, surrounding can also give heat to the system.

c, may be the process is adiabatic

d, may be adiabatic.

Q gas of certain mass at 273K is expanded to 81 times its vol. under adia. condition. if gamma = 1.25 for gas then its final temp. = ?

Q for a const. vol. gas thermometer one should fill gas at high temp & low pressure WHY ?

Q iF TEMP of uniform rod is slightly increase by del T its moment of inertia about 90* bisector increase BY = ? (ANS 2alpha I del T ).

Q Al sphere is dipped into H2O at 100*C if temp is increased forse of buoyancy will inc. OR decreased WHY =?

DENSITY OF A SUB. IS 10g/cc at 0*C &at 100*C d= 9.7 g/cc . coeff. of linear expantion is =?

**Asked By: SARIKA**

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**Joshi sir comment**

Answer : For adiabatic process TV^{γ-1 }= K

so 273*V^{0.25 }= T*(81V)^{0.25}

so 273/3 = T

Answer : Because real gases perform ideal behaviour in this range.

Answer : Inertia about perpendicular bisector = 1/12 ML^{2}

on increasing temperature by dT new inertia = 1/12 M[L(1+αdT)]^{2}

so increment in inertia = 1/12 ML^{2} [(1+αdT)^{2}-1^{-}] = 1/12 ML^{2} [ (αdT)^{2} + 2αdT ] now neglect (αdT)^{2 }because it is small term

Answer : Since temperature is already 100 degree centigrade so on increasing it further density of Al will increase rapidly in comparision to density of water so Buoyancy force will be increased.

Answer : use the formula d_{100} = d_{0 }/ (1+γ100) here γ=3α calculate α

A copper cube of each side 40 cm floats on mercury. the density of copper cube is 3.2g/cc and of mercury is 13.6g/cc at 27° C. the coefficient of volume expansion of Hg and linear expansion of copper are 1.8 * 10^(-4) /°C and 3 * 10^(-6) /°C respectively. Calculate the increase in height i.e. how much block sink further when the temperature rises from 27°C to 100°C?

**Asked By: HIMANSHU MITTAL**

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**Joshi sir comment**

dipped length of the cube in first case = m/Ad here m, A, d are mass of cube, area of the cube and density of mercury respectively

after increasing temperature dipped length in the second case = m/A(1+2αt)[d/(1+γt)]

values are m = VD = 40^{3} 3.2, A = 40^{2}, α = 3 * 10^{-6}, d = 13.6, γ = 1.8 * 10^{-4}, t = 73

all values are in CGS

now solve then substract 1st length from 2nd length