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a fixed amount of dry air at temperature of 27degree celsius is compressed to 1/9 of its original volume.its final temperature is?(gamma=1.5) a.627degree celsius b.600degree celsius c.158 degree celius d.527 degree celsius plz explain with solution

Answer :

kvnmurty
we treat dry air as an ideal gas.

Volume = V1          T1 = 300 Kelvin        Pressure = P1
V2 = V1 / 9              γ = ratio of specific heats = 1.5      T2 = ?    P2 = ?
number of moles  n is fixed.

     P1 V1 / T1  = P2 V2 / T2
     P1 V1 / 300 =  P2 (V1 /9) / T2      =>     T2 =  33.33  P2 / P1

   Is the gas compressed adiabatically and/or suddenly ? with no energy exchanged between the system and the environment ?  It has to be mentioned whether it  was an isobaric compression or adiabatic compression.  Otherwise the final pressure is to be mentioned.

[tex]P V^{\gamma} = Constant\\V^{\gamma-1} T = constant\\V^{0.5}T=constant\\T=\frac{constant}{\sqrt{V}}\\\\\frac{T2}{T1}=\sqrt{\frac{V1}{V2}}[/tex]

   T2 = 300 * √9 = 900⁰K  = 627⁰ C


TPS
V₁ = V
V₂ = V/9
T₁ = 27⁰C = (27+273)K = 300K
T₂ = ?

[tex]T_1V_1^{\gamma -1}=T_2V_2^{\gamma -1}\\ \\ \Rightarrow T_2= \frac{T_1V_1^{\gamma -1}}{V_2^{\gamma -1}} \\ \\ \Rightarrow T_2=T_1( \frac{V_1}{V_2})^{\gamma -1}\\ \\ \Rightarrow T_2= 300 \times (\frac{V }{V/9})^{1.5-1}\\ \\ \Rightarrow T_2= 300 \times (9)^{0.5}\\ \\ \Rightarrow T_2= 300 \times 3=900K[/tex]

T₂ = 900K = 900-273 = 627°C
Answer is A.

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