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The work done in adiabatic compression of 2 moles of ideal monatomic gas against constant external pressureof  2 atm  starting from initial pressure of 1 atmosphere and initial temperature of 300 Kelvin given R=2 Cal/mole degree
Asked by deepakudgiri29 | 27 Jan, 2019, 11:44: AM
answered-by-expert Expert Answer
Initial volume Vi is obtained by gas equation : piVi = nRT
 
Vi = (2×2×4.2×300)/(1.01×105)  ≈ 5×10-2 m3
 
for adiabatic process piViγ = pfVfγ , where γ is ratio of specific heat. For ideal mono-atomic gas γ = 1.67;
 
hence Vf = Vi (pi / pf )1/γ = 5×10-2 × (1/2)1/1.67 = 3.3×10-2 m3
 
Workdone for adiabatic process = begin mathsize 12px style 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 equals space fraction numerator 1 over denominator 0.67 end fraction left parenthesis 2 cross times 3.3 space minus space 1 cross times 5 right parenthesis cross times 1.01 cross times 10 to the power of 5 cross times 10 to the power of negative 2 end exponent space almost equal to space 2412 space J end style
Answered by Thiyagarajan K | 27 Jan, 2019, 15:40: PM

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