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An ideal gas absorbs $210 \mathrm{~J}$ of heat and undergoes expansion from $3 \mathrm{~L}$ to $6 \mathrm{~L}$ against a constant external pressure of $10^5 \mathrm{~Pa}$. What is the value of $\Delta \mathrm{U}$ ?
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Verified Answer
The correct answer is:
$-90 \mathrm{~J}$
$1 \mathrm{~Pa}=1 \times 10^{-5}$ bar
$\therefore \quad 10^5 \mathrm{~Pa}=1 \mathrm{bar}$
$\mathrm{W}=-\mathrm{P}_{\mathrm{ext}} \times \Delta \mathrm{V}=-1 \times(6-3)=-3 \mathrm{dm}^3$ bar
$1 \mathrm{dm}^3 \mathrm{bar}=100 \mathrm{~J}$
$\therefore \quad \mathrm{W}=-300 \mathrm{~J}$
$\mathrm{Q}=+210 \mathrm{~J}$
According to first law of thermodynamics,
$\Delta \mathrm{U}=\mathrm{Q}+\mathrm{W}$
$\therefore \quad \Delta U=+210 \mathrm{~J}+(-300 \mathrm{~J})=-90 \mathrm{~J}$
$\therefore \quad 10^5 \mathrm{~Pa}=1 \mathrm{bar}$
$\mathrm{W}=-\mathrm{P}_{\mathrm{ext}} \times \Delta \mathrm{V}=-1 \times(6-3)=-3 \mathrm{dm}^3$ bar
$1 \mathrm{dm}^3 \mathrm{bar}=100 \mathrm{~J}$
$\therefore \quad \mathrm{W}=-300 \mathrm{~J}$
$\mathrm{Q}=+210 \mathrm{~J}$
According to first law of thermodynamics,
$\Delta \mathrm{U}=\mathrm{Q}+\mathrm{W}$
$\therefore \quad \Delta U=+210 \mathrm{~J}+(-300 \mathrm{~J})=-90 \mathrm{~J}$
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