16g of nitrogen gas at STP are adiabatically compressed to a pressure of 22 atm.
ID: 2164672 • Letter: 1
Question
16g of nitrogen gas at STP are adiabatically compressed to a pressure of 22 atm.Question: What is the work done on the gas?
Please explain in FULL; I found T_final to be 390 degrees C.
Explanation / Answer
An ideal gas undergoing an adiabatic, reversible process satisfies the condition: p·V^? = constant throughout the process. ? = Cp/Cv is the heat capacity ratio. For a diatomic ideal gas like nitrogen it is ? = (7/2)·R / (5/2)·R = 7/5 Using ideal gas you may rewrite adiabatic equation as: p^(?-1)·T^(-?) = constant Hence: p1^(?-1) / T1^? = p2^(?-1) / T2^? => T2 = T1 · (p2/p1)^[(?-1)/?] = 273.15K · (22atm / 1atm)^[(7/5-1)/(7/5)] = 273.15K · (22)^[2/7] = 660.62K The work done on the gas is given by the integral W = - ? p dV from V1 to V2 V can expressed in terms of V by adiabatic equation: p·V^? = constant = p1·V1^? => V = V1·p1^(1/?) ·p^(-1/?) => dV = V1·p1^(1/?) · (-1/?)·p^(-1/? - 1) dp => W = - ? p dV from V1 to V2 = - ? p · V1·p1^(1/?) · (-1/?)·p^(-1/? - 1) dp from p1 to p2 = V1·p1^(1/?) · (1/?) · ? p^(-1/?) dp from p1 to p2 = V1·p1^(1/?) · (1/?) · [1/(1 - 1/?)] · { p2^[(1 - 1/?)] - p1^[(1 - 1/?)] } = V1·p1^(1/?) · [1/(? - 1)] · { p2^[(1 - 1/?)] - p1^[(1 - 1/?)] } = V1·p1 · [1/(? - 1)] · { (p2/p1)^[(1 - 1/?)] - 1 } = n·R·T1 · [1/(? - 1)] · { (p2/p1)^[(1 - 1/?)] - 1 } For this process W = (16g/28g/mol) · 8.314472J/molK · 273.15K · (5/2) · { (22)^(2/7) -1} = 4602.24 J
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