0.040 g of He is kept in a closed container initially at $100 \cdot 0^{\circ} \mathrm{C}$. The conta

Question

0.040 g of He is kept in a closed container initially at $100 \cdot 0^{\circ} \mathrm{C}$. The container is now heated. Neglecting the expansion of the container, calculate the temperature at which the internal energy is increased by 12 J .

Professor PC Thomas Classes · Answer

Helium ( He ) is a monoatomic gas, so its molar heat capacity at constant volume is $C_v=\frac{3}{2} R$. The molar mass of Helium is $M=4 \mathrm{~g} / \mathrm{mol}$. First, calculate the number of moles ( $n$ ): 
$ n=\frac{m}{M}=\frac{0.040}{4}=0.01 \mathrm{~mol} $
The increase in internal energy ( $\Delta U$ ) for a gas at constant volume is given by: 
$ \Delta U=n C_v \Delta T=n\left(\frac{3}{2} Right) \Delta T $
Substitute the given values to find the change in temperature $(\Delta T)$ : 
$ \begin{gathered} 12=(0.01) 	imes\left(\frac{3}{2} 	imes 8.31ight) 	imes \Delta T \ 12=0.015 	imes 8.31 	imes \Delta T \ 12=0.12465 	imes \Delta T \ \Delta T=\frac{12}{0.12465} \approx 96.2^{\circ} \mathrm{C} \end{gathered} $
The final temperature $T_f$ is: 
$ \begin{gathered} T_f=T_i+\Delta T \ T_f=100.0^{\circ} \mathrm{C}+96.2^{\circ} \mathrm{C}=196.2^{\circ} \mathrm{C} \end{gathered} $
Rounding to the nearest integer given in the options, the final temperature is $196^{\circ} \mathrm{C}$.