The amount of heat energy required to raise t | vedclass

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(D) Since the volume of the gas remains constant,the heat energy required is given by $\Delta Q = n C_v \Delta T$.Here,the number of moles $n$ for $1\

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Since the volume of the gas remains constant,the heat energy required is given by $\Delta Q = n C_v \Delta T$.Here,the number of moles $n$ for $1\,g$ of Helium (molar mass $M = 4\,g/mol$) is $n = \frac{1}{4}$.For a monoatomic gas like Helium,the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.The change in temperature is $\Delta T = T_2 - T_1$.Substituting these values,we get $\Delta Q = \frac{1}{4} 	imes \left( \frac{3}{2} R ight) 	imes (T_2 - T_1) = \frac{3}{8} R (T_2 - T_1)$.Using the relation $R = N_a k_B$,where $N_a$ is Avogadro's number and $k_B$ is the Boltzmann constant,we get $\Delta Q = \frac{3}{8} N_a k_B (T_2 - T_1)$.Thus,the correct option is $D$.

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

Similar Questions

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

Statement $A: C_P - C_V = R$
Statement $B: \frac{C_P}{C_V} = 1.67$

The specific heat at constant volume for the monoatomic argon is $0.075 \, kcal/kg-K$,whereas its gram molecular specific heat $C_V = 2.98 \, cal/mole/K$. The mass of the argon atom is (Avogadro's number $= 6.02 \times 10^{23} \, molecules/mole$)

Derive the ratio of $\frac{C_{P}}{C_{V}}$ for a diatomic gas.

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

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