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70 calories of heat are required to raise the temperature of 2 moles of an ideal gas at constant pressure from $30^{\circ} \mathrm{C}$ to $35^{\circ} \mathrm{C}$.
The amount of heat required to raise the temperature of the same sample of the gas through the same range at constant volume is $($ Gas constant $=1.99 \mathrm{cal} / \mathrm{K}-$ mole $)$
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The amount of heat required to raise the temperature of the same sample of the gas through the same range at constant volume is $($ Gas constant $=1.99 \mathrm{cal} / \mathrm{K}-$ mole $)$
Solution:
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Verified Answer
The correct answer is:
$50 \mathrm{cal}$
As $d Q=C_P m \Delta T$
$\Rightarrow \quad 70=C_P \times 2(35-30)$
$\Rightarrow \quad C_P=7 \mathrm{cal} / \mathrm{mole} /{ }^{\circ} \mathrm{C}$
$C_V=C_P-R$
$=7-1.99=5.01 \mathrm{cal} \mathrm{mol}^{-1}{ }^{\circ} \mathrm{C}^{-1}$
$\therefore \quad d Q^{\prime}=C_V m \Delta \mathrm{T}$
$=5.01 \times 2 \times(35-30)=50.1 \mathrm{cal}$
$\Rightarrow \quad 70=C_P \times 2(35-30)$
$\Rightarrow \quad C_P=7 \mathrm{cal} / \mathrm{mole} /{ }^{\circ} \mathrm{C}$
$C_V=C_P-R$
$=7-1.99=5.01 \mathrm{cal} \mathrm{mol}^{-1}{ }^{\circ} \mathrm{C}^{-1}$
$\therefore \quad d Q^{\prime}=C_V m \Delta \mathrm{T}$
$=5.01 \times 2 \times(35-30)=50.1 \mathrm{cal}$
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