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$310 \mathrm{~J}$ of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from $25^{\circ} \mathrm{C}$ to $35^{\circ} \mathrm{C}$. The amount of heat required to raise the temperature of the gas through the same range at constant volume is
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The correct answer is:
$144 \mathrm{~J}$
At constant pressure,
Heat required $=n C_{p} \Delta T$
$\begin{array}{ll}\Rightarrow & 310=2 \times C_{p} \times(35-25) \\ \Rightarrow & C_{p}=\frac{310}{20}=15.5 \mathrm{~J} / \mathrm{mol} / \mathrm{K}\end{array}$
Similarly, at constant volume,
Heat required $=n C_{V} \Delta T$
$=2\left(C_{p}-R\right) \times(35-25)$
$\left[\because C_{p}-C_{V}=R\right]$
$=2 \times(15.5-8.3) \times 10$ $=2 \times 7.2 \times 10=144 \mathrm{~J}$
$=2 \times 7.2 \times 10=144 \mathrm{~J}$
Heat required $=n C_{p} \Delta T$
$\begin{array}{ll}\Rightarrow & 310=2 \times C_{p} \times(35-25) \\ \Rightarrow & C_{p}=\frac{310}{20}=15.5 \mathrm{~J} / \mathrm{mol} / \mathrm{K}\end{array}$
Similarly, at constant volume,
Heat required $=n C_{V} \Delta T$
$=2\left(C_{p}-R\right) \times(35-25)$
$\left[\because C_{p}-C_{V}=R\right]$
$=2 \times(15.5-8.3) \times 10$ $=2 \times 7.2 \times 10=144 \mathrm{~J}$
$=2 \times 7.2 \times 10=144 \mathrm{~J}$
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