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Home / Doubts and Solutions/NEET/Physics

The work of 146KJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7 degree C Identify the atomicity of the gas Given R=8.3J/mol/K.  Please explain why 7 Degree celcius is taken equal to 7Kelvin while solving

Asked by nipunverma59 23rd September 2018, 2:54 PM
Answered by Expert
Answer:
 
Workdone W in adiabatic process is given by, begin mathsize 12px style W space equals space fraction numerator 1 over denominator gamma minus 1 end fraction open parentheses p subscript f V subscript f space minus space p subscript i V subscript i close parentheses space................... left parenthesis 1 right parenthesis end style
where γ is ratio of specific heats, pi and pf are initial and final presuure, Vi and Vf are initial and final volume.
 
if we substitute ideal gas equation of state, pV = nRT, then eqn.(1) is written as, begin mathsize 12px style W space equals space fraction numerator n cross times R over denominator gamma minus 1 end fraction open parentheses T subscript f space minus space T subscript i close parentheses space................... left parenthesis 2 right parenthesis end style
where n is number of moles, R is molar gas constant, Ti and Tf are initial and final temperatures.
 
let us substitute given data in eqn.(2) to get γ , 146×1000 = (1000×8.3×7)/(γ-1) ................(3)
 
we get γ≈1.4 from eqn.(3). Hence the gas is diatomic gas. atomicity is 2.
 
In thermodynamics, temperature is always expressed in kelvin scale ( for formulas and calculations).
 
Kelvin scale is given by,   TK = TC + 273 , where TK is temperature in Kelvin and other one in celsius.
since they are linearly related (only linear displacement), difference of temperature is always same.
Answered by Expert 23rd September 2018, 7:28 PM
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