A gas is allowed to expand in a well insulated container against a constant external pressure of `2.5 atm` from an initial volume of `2.50 L` to a final volume of `4.50 L`. The change in internal energy `Delta U` of the gas in joules will be:

Options

A. `1136.25 J`
B. `-500 J`
C. `-505 J`
D. `+505 J`

Answer

C

Explanation

Key concept Accorgind to first law of thermodynamics,
`DeltaU=q+w`
where, `DeltaU` =internal energy
q=heat absorbed or evolved , w=work done .
Also , work done against constant external pressure (irreversible process.)
`w=-p_("ext")DeltaV`
Work done in irreversible process,
`w=-p_("ext")DeltaV=-p_("ext")(V_(2)-V_(1))`
`-25 atm (4.5-2.5 L)`
`-5 L atm =-5xx1013` J
`=-505 J`
Since the system is well insulated , q=0
`therefore DeltaU=w=-505 J`
Hence, change in internal energy, `DeltaU` of the gas is -505 J.