A sample of helium behaves as an ideal gas as energy is
addedby heat at constant pressure from 273K to 393 K. If 20.0 J of
work is done by the gas during thisprocess, what is the mass of
helium present?

Respuesta :

Answer:

0.0321 g

Explanation:

Let helium specific heat [tex]c_h = 5.193 J/g K[/tex]

Assuming no energy is lost in the process, by the law of energy conservation we can state that the 20J work done is from the heat transfer to heat it up from 273K to 393K, which is a difference of ΔT = 393 - 273 = 120 K. We have the following heat transfer equation:

[tex]E_h = m_hc_h \Delta T = 20 J[/tex]

where [tex]m_h[/tex] is the mass of helium, which we are looking for:

[tex] m_h = \frac{20}{c_h \Delta T} = \frac{20}{5.193 * 120} \approx 0.0321 g[/tex]