One mole of water, intially at 50 degrees celcius is heated to140 degrees celciu
ID: 690599 • Letter: O
Question
One mole of water, intially at 50 degrees celcius is heated to140 degrees celcius at a constant pressure of 1.00 atm. Calcualtethe work for this process, assume idea gas behavior. I know this is simple, w= -pext(change in V), but I'm gettingthe wrong answer. One mole of water, intially at 50 degrees celcius is heated to140 degrees celcius at a constant pressure of 1.00 atm. Calcualtethe work for this process, assume idea gas behavior. I know this is simple, w= -pext(change in V), but I'm gettingthe wrong answer.Explanation / Answer
We know PV = nRT P = pressure = 1 atm T = initialtemperature = 50 oC = 50 + 273 = 323 K n = No .of moles = 1 R = gas constant = 0.0821 L atm / mol - K V = Volume of the water at 50 oC = ? Plug the values we get initial Volume , V = nRT / P = 26.5 L Also PV' = nRT' P = pressure = 1 atm T = final temperature = 140 oC = 140 + 273 = 413 K n = No .of moles = 1 R = gas constant = 0.0821 L atm / mol - K V = Volume of the water at 140 oC = ? Plug the values we get initial Volume , V' = nRT' / P = 33.9 L Work done , W = P V =P * ( V' - V ) = 1 atm * ( 33.9 - 26.5 ) = 101325 Pa * 7.4 L = 101325 Pa * 7.4 * 10^-3m^3 Since 1 L = 10^-3 m^3 = 749805 * 10^-3 J = 749.805 J P = pressure = 1 atm T = final temperature = 140 oC = 140 + 273 = 413 K n = No .of moles = 1 R = gas constant = 0.0821 L atm / mol - K V = Volume of the water at 140 oC = ? Plug the values we get initial Volume , V' = nRT' / P = 33.9 L Work done , W = P V =P * ( V' - V ) = 1 atm * ( 33.9 - 26.5 ) = 101325 Pa * 7.4 L = 101325 Pa * 7.4 * 10^-3m^3 Since 1 L = 10^-3 m^3 = 749805 * 10^-3 J = 749.805 JRelated Questions
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