1 , mole of a gas requires 40 , calories of h | vedclass

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

Question

(A) Given: $\mu = 1 \, mole$,$\Delta T = 30^{\circ}C - 20^{\circ}C = 10 \, K$,$(\Delta Q)_p = 40 \, cal$,$R = 2 \, cal \, mol^{-1} K^{-1}$.At constant

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) Given: $\mu = 1 \, mole$,$\Delta T = 30^{\circ}C - 20^{\circ}C = 10 \, K$,$(\Delta Q)_p = 40 \, cal$,$R = 2 \, cal \, mol^{-1} K^{-1}$.At constant pressure,the heat supplied is given by $(\Delta Q)_p = \mu C_p \Delta T$.Substituting the values: $40 = 1 	imes C_p 	imes 10$,which gives $C_p = 4 \, cal \, mol^{-1} K^{-1}$.Using Mayer's relation,$C_p - C_v = R$,we find $C_v = C_p - R = 4 - 2 = 2 \, cal \, mol^{-1} K^{-1}$.At constant volume,the heat supplied is given by $(\Delta Q)_v = \mu C_v \Delta T$.Substituting the values: $(\Delta Q)_v = 1 	imes 2 	imes 10 = 20 \, calories$.

$1 \, mole$ of a gas requires $40 \, calories$ of heat to raise its temperature from $20^{\circ}C$ to $30^{\circ}C$ at constant pressure. How many calories of heat are required to raise the temperature of the same gas by the same amount at constant volume? $(R = 2 \, cal \, mol^{-1} K^{-1})$

Similar Questions

Define two specific heats of a gas. Write the relation between them.

The molar specific heat at constant pressure of an ideal gas is $(7/2)R$. The ratio of specific heat at constant pressure to that at constant volume is

According to the law of equipartition of energy, the molar specific heat of a diatomic gas at constant volume, where the molecule has one additional vibrational mode, is:

The specific heat at constant pressure $(C_P)$ is greater than the specific heat at constant volume $(C_V)$ for the same gas because:

For a monoatomic gas,the work done at constant pressure is $W$. For the same rise in temperature of the gas,the heat supplied at constant volume is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module