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Question
A gas is allowed to expand in a well insulated container against a constant external pressure of 2 atm from an initial volume of 4.25 L to a final volume of 6.25 L. The change in internal energy (ΔU) of the gas will be:
A gas is allowed to expand in a well insulated container against a constant external pressure of 2 atm from an initial volume of 4.25 L to a final volume of 6.25 L. The change in internal energy (ΔU) of the gas will be:
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Solution
The correct option is D – 405.2 J
w=−PextΔV=−2(6.25−4.25)
=−4 L atm
=−4×101.3 J [∵1 L atm=101.3 J]
=−405.2 J
From first law of thermodynamics,
ΔU=q+w
q=0 for adiabatic process
ΔU=w=−405.2 J
w=−PextΔV=−2(6.25−4.25)
=−4 L atm
=−4×101.3 J [∵1 L atm=101.3 J]
=−405.2 J
From first law of thermodynamics,
ΔU=q+w
q=0 for adiabatic process
ΔU=w=−405.2 J
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