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The value of $\Delta H$ for cooling 2 mole of an ideal monoatomic gas from $225^{\circ} \mathrm{C}$ to $125^{\circ} \mathrm{C}$ at constant pressure will be [given $\left.C_{p}=\frac{5}{2} R\right]$
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Verified Answer
The correct answer is:
$-500 R$
$\Delta H=-n C_{p} \Delta T=-2 \times \frac{5}{2} R(225-125)$
$$
=-5 R(100)=-500 R
$$
$$
=-5 R(100)=-500 R
$$
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