0.040 g of He is kept in a closed container initially at 100.ºC. The container is now heated. Neglecting the expansion of the container, calculate the temperature at which the internal energy is increased by

Engineering

Physics

SI Units

Degree of Freedom Internal Energy of Gas and Specific Heat Capacity of Gases

Thermodynamics

Question

0.040 g of He is kept in a closed container initially at 100.ºC. The container is now heated. Neglecting the expansion of the container, calculate the temperature at which the internal energy is increased by $[R = \frac{25}{3} j - {mol}^{- 1} - k^{- 1}]$

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The mass of the Helium m = 0.040 g

The initial temperature of the container is T = 100ºC

The molecular mass of the helium is M_He = 4 g

The internal energy of the gas molecules:

U = \frac{3}{2} nRt

= \frac{3}{2} \times \frac{m}{M} \times RT

According to the question

\frac{3}{2} \times \frac{m}{M} \times RT + 12 = \frac{3}{2} \times \frac{m}{M} \times RT

⇒ 1.5 × 0.01 × 8.3 × 373 = 12 = 1.5 × 0.01 × 8.3 × T

\Rightarrow T = \frac{58.4385}{0.1245}

= 469.3855 K = 196.3ºC ≈ 196ºC