n = 3.45 mol of Hydrogen gas is initially at T = 346 K temperature and pi = 3.13

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Change your value Question n = 2.17 mol of Hydrogen gas is initially at T = 323 K temperature and pi = 2.61×105 Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches pf = 7.85×105 Pa. What would be the temperature of the gas, if the gas was allowed to adiabatically expand back to its original pressure? Answer Let …… 1 = initial state …… 2 = state at the end of isothermal compression …… T2 = T1 …… 3 = state at the end of adiabatic expansion …… p3 = p1 …… T3 = ??? Assuming a perfect gas … p3 V3 = n R T3 ---> T3 = p3 V3 / [ n R ] = p1 V3 / [ n R ] …… find V3 …… use the equation for an adiabatic process …… p V ^ ? = constant …… that is …… p3 V3 ^ ? = p1 V3 ^ ? = p2 V2 ^ ? ---> V3 ^ ? = [ p2 / p1 ] V2 ^ ? … …… V3 = V2 [ p2 / p1 ] ^ ( 1 / ? ) …… where ? = Cp / Cv = [ 7/2 ] / [ 5/2 ] …… …… ? = 7 / 5 = 1.4 for a diatomic gas, like H2 gas …… find V2 …… …… p2 V2 = n R T2 = n R T1 ---> V2 = n R T1 / p2 …… that is …… …… V3 = V2 [ p2 / p1 ] ^ ( 1 / ? ) = [ n R T1 / p2 ] [ p2 / p1 ] ^ ( 1 / ? ) …… T3 = p1 V3 / [ n R ] = p1 { [ n R T1 / p2 ] [ p2 / p1 ] ^ ( 1 / ? ) } / [ n R ] ………. = T1 [ p1 / p2 ] [ p2 / p1 ] ^ ( 1 / ? ) = T1 [ p2 / p1 ] ^ [ ( 1 ? ? ) / ? ] ………. = [ 323 K ] [ (7.85 × 10 5 ) / (2.61 × 10 5 ) ] ^ [ ( 1 ? 1.4 ) / 1.4 ] = 235.8 …… T3 = 236 K

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