The amount of heat energy required to raise t | vedclass

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

Question

(A) Since the volume of the gas remains constant,the heat energy required is given by $\Delta Q = n C_V \Delta T$.For Helium $(He)$,the molar mass is

Vedclass · Accepted Answer

$\frac{3}{8}{N_a}{k_B}\left( {{T_2} - {T_1}} \right)$

Vedclass · Answer

(A) Since the volume of the gas remains constant,the heat energy required is given by $\Delta Q = n C_V \Delta T$.For Helium $(He)$,the molar mass is $4\, g/mol$. Therefore,the number of moles $n$ in $1\, g$ of Helium is $n = \frac{1}{4}$.Helium is a monatomic gas,so its molar heat capacity at constant volume is $C_V = \frac{3}{2} R$.Given the temperature change $\Delta T = T_2 - T_1$,we substitute these values into the formula:$\Delta Q = n C_V \Delta T = \left( \frac{1}{4} \right) \left( \frac{3}{2} R \right) (T_2 - T_1) = \frac{3}{8} R (T_2 - T_1)$.Using the relation $R = N_a k_B$,where $N_a$ is Avogadro's number and $k_B$ is the Boltzmann constant,we get:$\Delta Q = \frac{3}{8} N_a k_B (T_2 - T_1)$.

The amount of heat energy required to raise the temperature of $1\, g$ of Helium at $NTP$ from $T_1\, K$ to $T_2\, K$ is:

Similar Questions

$A$ diatomic gas molecule has translational,rotational,and vibrational degrees of freedom. The ratio ${C_P}/{C_V}$ is

The ratio of the specific heats $\frac{C_p}{C_v} = \gamma$ in terms of degrees of freedom $(n)$ is given by

At constant volume,the specific heat of a gas is $\frac{3R}{2}$. Then the value of $\gamma$ will be ....

The molar specific heat of an ideal gas at constant pressure and constant volume is $C_{p}$ and $C_{v}$ respectively. If $R$ is the universal gas constant and $\gamma = \frac{C_p}{C_v}$,then $C_v =$

Obtain the molar specific heat of a solid by using the law of equipartition of energy.

Vedclass Test Series

Exam Paper Generator

Online Exam Module