Heat and Thermodynamics 5 Question 57

58. An ideal gas having initial pressure p, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66V while its temperature falls to T/2.

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Answer:

Correct Answer: 58. 675K,3.6×106N/m2

Solution:

  1. (a) In adiabatic process TVγ1= constant

TVγ1=(T2)(5.66V)γ1 or (5.66)γ1=2(γ1)ln(5.66)=ln(2)γ1=0.4 or γ=1.4

i.e. degree of freedom, f=5 as γ=1+2f

(b) Using, pVγ= constant

pV1.4=pf(5.66V)1.4pf=0.09p

Now, work done in adiabatic process

W=piVipfVfγ1=(pV)(0.09p)(5.66V)1.41=1.23pV



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