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 Option 1) -500 J Option 2) -505 J Option 3) +505 J Option 4) 1136.25 J

As we learnt

Work Done for Irreversible Isothermal Expansion of an ideal gas -

$W= -P_{ext}\left ( V_{f} -V_{i}\right )$

- wherein

$P_{ext}$ may be equal to $P_{f}$ or may not be equal to $P_{f}$, but work done is always calculated by $P_{ext}$

and

Heat exchange between system and surrounding is zero i.e. $q= 0$

- wherein

$\Delta E= q+w$

$q= 0$

$\Delta E=w$

Work done in this process will be

W = $-P_{ext}^{\Delta V}$

= -2.5[4.5 - 2.5] = -5 L atm

= - 5 101.3 J = -206J

since system is well insuvated

q = 0

=>$\Delta U$_W = 0

or $\Delta U$ = +W = -505 J

Option 1)

-500 J

this is the incorrect option

Option 2)

-505 J

this is the correct option

Option 3)

+505 J

this is the incorrect option

Option 4)

1136.25 J

this is the incorrect option

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