Question

2L of an ideal gas at pressure of 10 atm expands isothermally into a vacuum until its total volume is 10L. How much work is done in the experiment?
A.8 L-atm
B.16.1 L-atm
C.24.0 L-atm
D.No work done

Answer 1

Hint: For isothermal processes the change in internal energy i.e.$\Delta \mu {\text{ or }}\Delta {\rm E}$ is zero.
We know that, for isothermal$\Delta {\rm T} = 0$
Also, $\Delta \mu = nCv\Delta T$
Therefore,$\Delta \mu = 0$

Complete step by step answer:
$\Delta \mu = Q + \omega $
$\Delta \mu $$ \to $Change in internal energy
$Q$$ \to $Heat added
$\omega $$ \to $Work done on the system
The first law of thermodynamics says that energy can be changed from one form to another, but it cannot be created or destroyed. The total amount of energy and the matter in the universe remains constant, merely changing from one form to another. The first law of thermodynamics (conservation) states that energy is always conserved, it cannot be created or destroyed. In essence, the energy can be converted from one form to another.
$\Delta \mu $=0 (isothermal process)
Also work done ($\omega $)= $ - {P_{ext}}\Delta V$
$ - {P_{ext}}$$ \to $External pressure
$\Delta V$$ \to $Change in volume
So as the question is telling, we are dealing with the vacuum state so
${P_{ext}} = 0$
$\omega $=$ - {P_{ext}}\Delta V$
So the work done= 0
According to the equation(1)
$\Delta \mu $=q+$\omega $ ($\Delta \mu $=0,$\omega $=0)
0= q+$\omega $
q= -$\omega $=0
So the work done is zero and so is there heat absorbed.
So the correct option is (D) i.e. no work done.

Note:
In the isothermal process ($\Delta {\rm T} = 0$) the change in internal energy ($\Delta \mu $) and the enthalpy change ($\Delta {\rm H}$) become zero.
 For the Isochoric process ($\Delta V = 0$), work done becomes zero.
Vacuum implies that the external pressure on the system is zero.