Thermodynamics

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Question
CBSEENPH11020644

4.0 g of a gas occupies 22.4 L at NTP. The specific heat capacity of the gas at constant volume is 5.0 JK-1 mol-1. If the speed of sound in this gas at NTP is 952 ms-1, then the heat capacity at constant pressure is.
(Take gas constant R = 8.3 Jk-1 mol-1)

  • 8.0 JK-1mol-1

  • 7.5 JK-1mol-1

  • 7.0 JK-1mol-1

  • 8.5 JK-1mol-1

Solution

A.

8.0 JK-1mol-1

Given, M = 4 gm
V= 22.4 L, 
CV = 5 JK-1 mol-1
straight v subscript sound space equals 952 space straight m divided by straight s comma
straight C subscript straight p space equals ?
As comma space velocity space of space sound comma space straight v subscript sound space equals space square root of γpV over straight M end root
rightwards double arrow space straight gamma space equals space straight M over pV straight V subscript sound superscript 2 space equals space straight C subscript straight p over straight C subscript straight v
so comma space heat space capacity space at space constant space pressure comma

straight C subscript straight p space equals space straight C subscript straight v open square brackets straight M over pV close square brackets straight v subscript sound superscript 2 space equals space 5 space open square brackets fraction numerator 4 space straight x space 10 to the power of negative 3 end exponent over denominator 10 to the power of 5 space straight x space 22.4 space straight x space 10 to the power of negative 3 end exponent end fraction close square brackets left parenthesis 952 right parenthesis squared

equals fraction numerator 20 over denominator 22.4 end fraction space straight x space left parenthesis 952 right parenthesis squared space straight x space 10 to the power of negative 5 end exponent

equals space 809.200 space straight x space 10 to the power of negative 5 end exponent space equals space 8.09 space straight j divided by mol space straight K.

Sponsor Area

Question
CBSEENPH11020787

A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency?

  • 275 K

  • 175 K

  • 250 K

  • 225 K

Solution

C.

250 K

The efficiency of Carnot engine is defined as the ratio of work done to the heat supplied i.e., 
                          straight eta space equals space fraction numerator Work space done over denominator Heat space supplied end fraction space equals straight W over straight Q subscript 1 space equals space fraction numerator straight Q subscript 1 minus straight Q subscript 2 over denominator straight Q subscript 1 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 minus straight Q subscript 2 over straight Q subscript 1 space equals space 1 minus straight T subscript 2 over straight T subscript 1
Here, straight T subscript 1 is the temperature of source and T2 is the temperature of sink
    As given,             straight eta space equals space 40 percent sign space equals space 40 over 100 space equals 0.4
and                straight T subscript 2 space equals space 300 space straight K
So,        
       0.4 space equals space 1 minus 300 over straight T subscript 1
rightwards double arrow space space space space straight T subscript 1 space equals space fraction numerator 300 over denominator 1 minus 0.4 end fraction equals fraction numerator 300 over denominator 0.6 end fraction space equals 500 space straight K
Let temperature of the source be increased by x K, then efficiency becomes
straight eta apostrophe space equals space 40 percent sign space plus space 50 percent sign space of space straight eta
space space space space space equals space 40 over 100 plus 50 over 100 cross times 0.4
space space space space space space equals 0.4 space plus space 0.5 space cross times space 0.4
space space space space space space space equals space 0.6
Hence comma space space space space space 0.6 space equals space 1 minus fraction numerator 300 over denominator 500 plus straight x end fraction
rightwards double arrow space space space space space space fraction numerator 300 over denominator 500 plus straight x end fraction space equals space 0.4
rightwards double arrow space space space space space space 500 plus straight x space equals space fraction numerator 300 over denominator 0.4 end fraction space equals space 750
therefore space space space space straight x space equals space 750 minus 500 space equals space 250 space straight K

Question
CBSEENPH12039768

A Carnot engine, having an efficiency of asstraight eta space equals 1 over 10 heat engine, is used as a refrigerator. If the work is done on the system is 10J, the amount of energy absorbed from the reservoir at lower temperature is

  • 100J 

  • 99 J 

  • 90 J

  • 1 J 

Solution

C.

90 J

According to Carnot engine,
As Q1 +W = Q2
we space have space given comma
straight eta space equals space 1 over 10
Now comma space using space straight eta space equals space 1 minus straight T subscript 1 over straight T subscript 2
So comma space 1 over 10 space equals 1 minus space straight T subscript 1 over straight T subscript 2
rightwards double arrow space straight T subscript 1 over straight T subscript 2 space equals space 9 over 10
Now space straight Q subscript 1 over straight Q subscript 2 space equals space straight T subscript 1 over straight T subscript 2
space rightwards double arrow space fraction numerator straight Q subscript 1 over denominator straight Q subscript 1 plus space straight W end fraction space equals space 9 over 10
10 straight Q subscript 1 space equals space 9 straight Q subscript 1 plus 9 straight W
space straight Q subscript 1 space equals 9 straight W space equals space 9 space straight x space 10 space equals space 90 space straight J

Question
CBSEENPH11020519

A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced its half. Then,

  • compressing the gas through adiabatic process will require more work to b done

  • compressing the gas through isothermally or adiabatically will require the same amount of work.

  • which of the case (whether compression through isohermal or through adiabatic process) requires more work will depend upon the atomicity of the gas.

  • compressing the gas isothermally will require more work to be done

Solution

B.

compressing the gas through isothermally or adiabatically will require the same amount of work.

Plotting P-V graph for the compression of a gas isothermally and adiabatically simultaneously to half of its initial volume.

An isothermal curve is less steeper than the adiabatic curve. So, area under the P-V curve for adiabatic process has more magnitude than the isothermal curve. Hence, work done in adiabatic process will be more than in isothermal process.

Question
CBSEENPH11020615

A gas is taken through the cycle A → B → C → A, as shown, What is the net work done by the gas?


  • 2000 J

  • 1000 J

  • Zero

  • -2000 J 

Solution

B.

1000 J

Net work done = Area enclosed in pV curve i.e. Δ ABC
 = 1 half space straight x space 5 space straight x space 10 to the power of negative 3 end exponent space straight x space 4 space straight x space 10 to the power of 5 space straight J = 1000 J

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