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

Answer 1

The molar-specific heat of an ideal gas at constant pressure and volume are C_p​ and C_v​ respectively.

1. 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)°C
• n= Given moles of substance= $2$ mole
• ${\mathrm{C}}_{\mathrm{p}}=\frac{70}{2\times (30-25)}=\frac{70}{2\times 5}=\frac{70}{10}=7{\mathrm{caloriesmole}}^{-1}{\mathrm{K}}^{-1}$
• For an ideal gas ${\mathrm{C}}_{\mathrm{p}}-{\mathrm{C}}_{\mathrm{v}}=\mathrm{R}=8.314\frac{\mathrm{J}}{\mathrm{Mole}\times \mathrm{K}}\simeq 2\frac{\mathrm{Cal}}{\mathrm{Mole}\times \mathrm{K}}$
• ${\mathrm{C}}_{\mathrm{v}}={\mathrm{C}}_{\mathrm{p}}-\mathrm{R}$
• ${\mathrm{C}}_{\mathrm{v}}=7-2=5{\mathrm{caloriesmole}}^{-1}{\mathrm{K}}^{-1}$

2. The molar-specific heat of an ideal gas for the isochoric process(at constant volume)

• $$\begin{gathered} {\mathrm{C}}_{\mathrm{v}}=\frac{∆\mathrm{Q}}{\mathrm{n}∆\mathrm{T}} \\ ∆\mathrm{Q}={\mathrm{C}}_{\mathrm{v}}\times \mathrm{n}∆\mathrm{T} \end{gathered}$$
• $$\begin{gathered} ∆\mathrm{Q}=5\times 2\times (30-25) \\ ∆\mathrm{Q}=5\times 2\times 5 \\ ∆\mathrm{Q}=5\times 10 \\ ∆\mathrm{Q}=50\mathrm{calories} \end{gathered}$$

$50$ $\mathrm{calories}$ heat is required to raise the temperature of $2$ Moles of an Ideal gas at a constant volume from $25°\mathrm{C}$ to $30°\mathrm{C}$.