1. n = 2.11 mol of Hydrogen gas is initially at T = 356 K temperature and pi = 2
ID: 2101265 • Letter: 1
Question
1. n = 2.11 mol of Hydrogen gas is initially at T = 356 K temperature and pi = 2.49Explanation / Answer
1) Volume of the gas PfVf = nRT Vf = nRT / Pf = 2.11 * 8.314*356 / 6.61*10^5 = 9.44*10^-3 m^3 = 0.0094 m^3 2) Vi = nRT / Pi = 2.11 * 8.314*356 / 2.49*10^5 = 0.0250 m^3 Work = nRT ln (Vf/Vi) = 6108 J 5) What would be the temperature of the gas, if the gas was allowed to adiabatically expand back to its original pressure 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 ? ? ) / ? ] = [ 356 K ] [ (6.61 × 10 5 ) / (2.49 × 10 5 ) ] ^ [ ( 1 - 1.4 ) / 1.4 ] =269.34 T3 = 269.34 K temperature of the gas, if the gas was allowed to adiabatically expand back to its original pressure is 217.95 K
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