Question

0.040 g of He is kept in a closed container initially at 100.0°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.

Answer 1

$$\begin{gathered} \begin{array}{l}\text{Here,}\\ m=0.040g\\ M=4g\\ n=\frac{0.040}{4}=0.01\\ T_1=\left(100+273\right)K=373K\\ \text{He is a monoatomic gas. Thus,}\\ C_v=3\times \left(\frac{1}{2}R\right)\\ \Rightarrow C_v=1.5\times 8.3=12.45\\ {\text{Let the initial internal energy be U}}_1\text{.}\\ {\text{Let the final internal energy be U}}_2\text{.}\\ U_2-U_1=n{C}_v(T_2-T_1)\\ \Rightarrow 0.01\times 12.45(T_2-373)=12\\ \Rightarrow T_2=469K\\ {\text{The temperature in}}^{0}C\text{can be obtained as follows:}\end{array} \\ {\mathrm{469\; -\; 273\; =\; 196}}^{0}C \end{gathered}$$