Suppose 1.90 mol of an ideal gasundergoes a reversible isothermal expansion from

Question

Suppose 1.90 mol of an ideal gasundergoes a reversible isothermal expansion from volumeV1 to volume V2 =2.00V1 at temperature T = 350 K. (a) Find the work done by the gas.
1 J

(b) Find the entropy change of the gas.
2 J/K

(c) If the expansion is reversible and adiabatic instead ofisothermal, what is the entropy change of the gas?
3 J/K (a) Find the work done by the gas.
1 J

(b) Find the entropy change of the gas.
2 J/K

(c) If the expansion is reversible and adiabatic instead ofisothermal, what is the entropy change of the gas?
3 J/K

Explanation / Answer

Initial volume V = V fina; volume V ' = 2 V temp T = 350 K No.of moes n = 1.9 mol (a). work done by the gas W = nRT ln ( V ' / V ) where R = gas constant = 8.314 J / mol K plug the values   W = 3832.27 J (b). entropy change = heat / T In isothermal process heat = work So, entropy change = 3832.27 / 350                                =10.94 J / K (c). entropy change = 0

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