Question

# $70$Cal heat is required to raise the temperature of $2$ moles of an Ideal gas at constant pressure from $25°\mathrm{C}$ to $30°\mathrm{C}$, then the amount of heat, required to raise the temperature of the same gas through the same rise of temp, at constant volume is

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Solution

## The molar-specific heat of an ideal gas at constant pressure and volume are Cp​ and Cv​ respectively.The molar-specific heat of an ideal gas for the isobaric process(at constant pressure)${\mathrm{C}}_{\mathrm{p}}=\frac{∆\mathrm{Q}}{\mathrm{n}∆\mathrm{T}}$Where $∆\mathrm{Q}$=Change in heat energy=70 Cal $∆\mathrm{T}$= Change in temperature=(30-25)°Cn= Given moles of substance= $2$ mole${\mathrm{C}}_{\mathrm{p}}=\frac{70}{2×\left(30-25\right)}=\frac{70}{2×5}=\frac{70}{10}=7{\mathrm{caloriesmole}}^{-1}{\mathrm{K}}^{-1}$For an ideal gas ${\mathrm{C}}_{\mathrm{p}}-{\mathrm{C}}_{\mathrm{v}}=\mathrm{R}=8.314\frac{\mathrm{J}}{\mathrm{Mole}×\mathrm{K}}\simeq 2\frac{\mathrm{Cal}}{\mathrm{Mole}×\mathrm{K}}$${\mathrm{C}}_{\mathrm{v}}={\mathrm{C}}_{\mathrm{p}}-\mathrm{R}$${\mathrm{C}}_{\mathrm{v}}=7-2=5{\mathrm{caloriesmole}}^{-1}{\mathrm{K}}^{-1}$ 2. The molar-specific heat of an ideal gas for the isochoric process(at constant volume)${\mathrm{C}}_{\mathrm{v}}=\frac{∆\mathrm{Q}}{\mathrm{n}∆\mathrm{T}}\phantom{\rule{0ex}{0ex}}∆\mathrm{Q}={\mathrm{C}}_{\mathrm{v}}×\mathrm{n}∆\mathrm{T}$$∆\mathrm{Q}=5×2×\left(30-25\right)\phantom{\rule{0ex}{0ex}}∆\mathrm{Q}=5×2×5\phantom{\rule{0ex}{0ex}}∆\mathrm{Q}=5×10\phantom{\rule{0ex}{0ex}}∆\mathrm{Q}=50\mathrm{calories}\phantom{\rule{0ex}{0ex}}$ $50$ $\mathrm{calories}$ heat is required to raise the temperature of $2$ Moles of an Ideal gas at a constant volume from $25°\mathrm{C}$ to $30°\mathrm{C}$.

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