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1 mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C.  The work done is 3 kJ. The final temperature (K) of the gas is equal to (CV = 20 J mol–1K–1):
    Correct answer is '150'. Can you explain this answer?
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    1 mole of an ideal gas is allowed to expand reversibly and adiabatical...
    answer is correct 
    like in carnot cycle when we go for expansion adiabatically nd reversibly then temp always dec. i.e. T(H) to T(L)
    so in this also temp goes on decrease coz on expansion work done should be negative then 
    dT = 150 is literally as 
    T(L)-T(H) = -150 
    so T(L) should be 150 to make work done negative

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    1 mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. The work done is 3 kJ. The final temperature (K) of the gas is equal to (CV = 20 J mol–1K–1):Correct answer is '150'. Can you explain this answer?
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