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.

Use R = 8.3 J K^-1 mol^-1

Answer 1

Here ,
m = 0.040g
M = 4g
n = \[\frac{0.040}{4} = 0.01 \]
T₁ = (100 + 273) K = 373 K
He is a monoatomic gas. Thus,
\[\ C_v = 3 × (\frac{1}{2}R) \] 
⇒ C_v = 1.5 × 8.3 = 12.45
Let the initial internal energy be U₁ .
Let the final internal energy be U₂ .
U₂ -U₁ = n C_v ( T₂ -T₁ )
⇒ 0.01 × 12.45( T₂ - 373) = 12
⇒ T₂ = 469  K
The temperature in °C can be obtained as follows: 469 - 273 = 196° C