40 , text{calories} of heat is needed to rais | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For an ideal monoatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$ and at constant pressure is $C_p = \frac{5}{2}R$.Gi

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) For an ideal monoatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$ and at constant pressure is $C_p = \frac{5}{2}R$.Given $R = 2 \, 	ext{cal} \, 	ext{mol}^{-1} 	ext{K}^{-1}$,we have $C_v = \frac{3}{2} 	imes 2 = 3 \, 	ext{cal} \, 	ext{mol}^{-1} 	ext{K}^{-1}$ and $C_p = \frac{5}{2} 	imes 2 = 5 \, 	ext{cal} \, 	ext{mol}^{-1} 	ext{K}^{-1}$.For a constant pressure process,the heat supplied is $Q_p = n C_p \Delta T$.Given $Q_p = 40 \, 	ext{cal}$,$n = 1 \, 	ext{mole}$,and $\Delta T = 30^{\circ}C - 20^{\circ}C = 10 \, 	ext{K}$.Checking the consistency: $Q_p = 1 	imes 5 	imes 10 = 50 \, 	ext{cal}$.However,the problem states $40 \, 	ext{cal}$ is required. This implies the gas might not be perfectly monoatomic or the provided $R$ is for calculation purposes. Using the ratio $C_v/C_p = 1/\gamma$ where $\gamma = 5/3$ for monoatomic gas:$C_v = C_p / \gamma = (C_p 	imes 3) / 5$.From $Q_p = n C_p \Delta T$,we have $40 = 1 	imes C_p 	imes 10$,so $C_p = 4 \, 	ext{cal} \, 	ext{mol}^{-1} 	ext{K}^{-1}$.Then $C_v = 4 	imes (3/5) = 2.4 \, 	ext{cal} \, 	ext{mol}^{-1} 	ext{K}^{-1}$.For constant volume,$Q_v = n C_v \Delta T = 1 	imes 2.4 	imes 10 = 24 \, 	ext{cal}$.

Similar Questions

For hydrogen gas,$C_P - C_V = a$,and for oxygen gas,$C_P - C_V = b$. The relationship between $a$ and $b$ is given by: (where $C_P$ and $C_V$ are molar specific heats at constant pressure and constant volume,respectively.)

Write the value of $\frac{C_P}{C_V}$ for a monoatomic gas.

The amount of heat required to raise the temperature of $2$ moles of helium gas from $0^{\circ}C$ to $100^{\circ}C$ at constant volume and constant pressure,respectively,is:

Let $\gamma_1$ be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and $\gamma_2$ be the similar ratio of a diatomic gas. Considering the diatomic gas molecule as a rigid rotator,the ratio $\frac{\gamma_1}{\gamma_2}$ is

For a gas,$R/C_V = 0.67$. What kind of molecules is this gas made of?

Vedclass Test Series

Exam Paper Generator

Online Exam Module