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Mechanical Properties Of Fluids

Question
CBSEENPH11020670

A mass of diatomic gas left parenthesis straight gamma space equals space 1.4 right parenthesis at a pressure of 2 atm is compressed adiabatically so that its temperature rise from 27oC to 927o C. The pressure of the gas is final state is 

  • 28 atm

  • 68.7 atm

  • 256 atm

  • 8 atm

Solution

C.

256 atm

T1 = 273 + 27 = 300k
T2 = 273+927 = 1200 K
Gas equation for adiabatic process
pV to the power of straight gamma space equals space constant

straight p open parentheses straight T over straight P close parentheses to the power of straight gamma space equals space constant
space
left parenthesis therefore space pV space equals space RT right parenthesis

therefore comma
straight p subscript 2 over straight p subscript 1 space equals space open parentheses straight T subscript 2 over straight T subscript 1 close parentheses to the power of fraction numerator straight gamma over denominator straight gamma minus 1 end fraction end exponent

Or

straight p subscript 2 space equals straight p subscript 1 space open parentheses straight T subscript 2 over straight T subscript 1 close parentheses to the power of fraction numerator straight gamma over denominator straight gamma minus 1 end fraction end exponent

straight p subscript 2 space equals space 2 open parentheses 1200 over 300 close parentheses to the power of fraction numerator 1.4 over denominator 1.4 minus 1 end fraction end exponent

straight p subscript 2 space equals space 256 space atm