Getting back to the old things #1

One mole of a mono atomic ideal gas undergoes a thermodynamic process such that V 3 / T 2 V^3/T^2 = constant. Then:

  1. Work done by gas is 200R when temperature of gas is raised by 300K.
  2. Specific heat of gas in the process is 13 R / 6 13R/6 .
  3. Work done by gas is 300R when temperature of gas is raised by 600K.
  4. The specific heat of the gas in the process is 21 R / 4 21R/4 .

Which of the above statements are true? Give the answer as the product of serial numbers of the right options. For ex. If 1 and 4 are correct the answer should be 1×4=4.


The answer is 2.

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1 solution

Spandan Senapati
May 1, 2017

For a polytropic process P V x = k PV^x=k the heat capacity is C = C ( v ) + R / ( 1 x ) C=C(v)+R/(1-x) .Here x = 1 / 2 x=-1/2 .So C = 13 R / 6 C=13R/6 .And similarly work done is n R ( T 2 T 1 ) / ( 1 x ) = W nR(T2-T1)/(1-x)=W .Clearly first 2 are correct.

Same solution

Md Zuhair - 3 years, 8 months ago

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