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

The temperature at which the internal energy is increased by 12 J is $1960 \mathrm{C}$.

Explanation:

Given data,  

$\mathrm{m}=0.040 \mathrm{g}$

$M=4 g$

$n=\frac{0.040}{4}=0.01$

$T_{1}=100+273 \mathrm{K}=373 \mathrm{K}$

He is a gas which is monoatomic. Accordingly,

$C_{v}=3 \times(12 \mathrm{R})$

$C_{v}=1.5 \times 8.3=12.45$

Let the primarily  internal energy be $U_{1}$.

Let the finally internal energy be $\mathrm{U}_{2}$.

$\mathrm{U}_{2}-\mathrm{U}_{1}=\mathrm{nC}_{\mathrm{V}}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)$

$0.01 \times 12.45\left(T_{2}-373\right)=12$

$T_{2}=469 \mathrm{K}$

You can get temperature in $^{0} \mathrm{C}$ as follows :-  

$469-273=1960 C$

Answer 2

The temperature at which the internal energy increased by 12 J is 196° C

Explanation:

Given data in the question  

Mass of Helium, m = 0.040 g

M = 4 g

$\begin{aligned}&n=\frac{0.040}{4}=0.01\\&T_{1}=100+273 \mathrm{K}=373 \mathrm{K}\end{aligned}$

He is a gas which is mono atomic. Accordingly,

$\begin{aligned}&\mathrm{C}_{4}=3 \times(12 \mathrm{R})\\&\mathrm{S}_{4}=1.5 \times 8.3=12.45\end{aligned}$

Let the primarily  internal energy be $U_1$.

Let the finally internal energy be $U_2$ .

$\begin{aligned}&\mathrm{U}_{2}-\mathrm{U}_{1}=\mathrm{nC}_{2}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)\\&0.01 \times 12.45\left(T_{2}-373\right)=12\end{aligned}$

$T_2 =469 K$

You can get temperature in °C as follows :-  

We have to subtract 273 from the temperature in Kelvin to convert in degree Celsius

469 – 273 = 196° C

Therefore the temperature of 0.040 g of helium at which the internal energy is increased by 12 J is 196°C when the initial temperature is 100° C