ThermodynamicsMedium
Question
one mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27oC. if the work done during the process is 3kJ, then final temperature of the gas is (Cv = 20J/K)
Options
A.100 K
B.150 K
C.195 K
D.255 K
Solution
Since the gas expands adiabatically (i.e., no change in enthalpy) so the heat is totallyconverted in to work
For the gas, Cv = 20J/K. Thes 20 J of heat is required for 1o change in temperature of the gas.Heat change involved during during the process i.e., work done = 3 kJ = 3000J.
Change in temperature =
K = 150K
Initial temprature = 300 K
Since, the gas thus final temperrature decreases and thus final tempeerature is300 - 150 = 150
For the gas, Cv = 20J/K. Thes 20 J of heat is required for 1o change in temperature of the gas.Heat change involved during during the process i.e., work done = 3 kJ = 3000J.
Change in temperature =
K = 150KInitial temprature = 300 K
Since, the gas thus final temperrature decreases and thus final tempeerature is300 - 150 = 150
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