1 mole of monoatomic ideal gas expands adiabatically at initial temperature 27° Celsius against a constant external pressure of 0.5 bar from 2 litre to 5 litre the change in enthalpy of process is
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0
Answer:
Explanation:
Work done against constant extemal pressure =P
ext
(V
2
−V
1
)
In adiabatic condition Δq=0
therefore w=Δu
−P
ext
(V
2
−V
1
)=
2
3
R(T
2
−T
1
) [Expansion work is negative]
On solving, T
2
=T
1
−
3×0.0821
2
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Answer:
step by step
plus and multiple take answer and divide 5
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