InterviewSolution
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InterviewSolution
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A certain volume of dry air at NTP is allowed to expand 4 times of its original volume under (a) isothermal conditions, (b) adiabatic conditions. Calculate the final pressure and temperature in each case if γ = 1.4. |
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Answer» Given : Suppose V1 = V V2 = 4V P1 = 76 cm of Hg P2 = ? γ = 1.4 T1 = 273K T2 = ? (a) For isothermal expansion : P1V1 = P2V2 or, P2 = \(P_1\frac{V_1}{V_2}\) = 19 cm of Hg. As the process is isothermal, therefore the final temperature will be the same as the initial temperature, i.e., T2 = 273 K (b) Adiabatic expansion : From the relation P1V1γ = P2V2γ ∴ P2 = P1\((\frac{V_1}{V_2})^γ\) \(= 76\times(\frac{1}{4})^{0.04}\) \(= 76\times(0.25)^{1.4}\) = 10.91 cm of Hg From the relation, T1V1γ−1 = T2V2γ−1 ⇒ T2 = T1\((\frac{V_1}{V_2})^{γ-1}\) ⇒ T2 = 273\((\frac{1}{4})^{0.04}\) = \(\frac{273}{(4)^{0.04}}\) = 156.8 K |
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