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  • In an isothermal process constant temperature process , 150 Joules of work is done on an ideal gas , calculate 1. Q and 2. delta U .

In an isothermal process constant temperature process , 150 Joules of work is done on an ideal gas , calculate 1. Q and 2. delta U .

Answers (1)

As the process is isothermal, thus the change in internal energy will be zero, i.e.,

ΔU=0

Now from first law of thermodynamics,

ΔU=q+w

?ΔU=0

∴q=−w

Given that, the work is done on the gas.

∴w=+150J

Therefore,

q=−w=−150J

Posted by

Satyajeet Kumar

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