fosilifer0ag

2023-03-14

Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 liter of ideal gas expands isothermally into vacuum unit its total volume is 5 litre?

unieventos8l9

Beginner2023-03-15Added 5 answers

Work done in vacuum is calculated by:

$w=-{P}_{\text{ext}}({V}_{\text{final}}-{V}_{\text{initial}})$

For the given problem: ${P}_{\text{ext}}=0$, because it is a free expansion (in the vacuum).

Hence,

$w=-(0\times (5-1))\phantom{\rule{0ex}{0ex}}w=0$

According to joule experimental observation, For an isothermal expansion of ideal gas, no heat will be absorbed or evolved, hence q=0.

By the 1st law of thermodynamics,

$q+w=\mathrm{\u25b3}U\phantom{\rule{0ex}{0ex}}\mathrm{\u25b3}U=0$

Since, w=0 and q=0, then the change in internal energy $\mathrm{\u25b3}U$ will also be 0.

$w=-{P}_{\text{ext}}({V}_{\text{final}}-{V}_{\text{initial}})$

For the given problem: ${P}_{\text{ext}}=0$, because it is a free expansion (in the vacuum).

Hence,

$w=-(0\times (5-1))\phantom{\rule{0ex}{0ex}}w=0$

According to joule experimental observation, For an isothermal expansion of ideal gas, no heat will be absorbed or evolved, hence q=0.

By the 1st law of thermodynamics,

$q+w=\mathrm{\u25b3}U\phantom{\rule{0ex}{0ex}}\mathrm{\u25b3}U=0$

Since, w=0 and q=0, then the change in internal energy $\mathrm{\u25b3}U$ will also be 0.