Question

Expansion of a gas in vacuum is called free expansion. Calculate the work done and change in internal energy when 1 mol of an ideal gas expands isothermally from 1L to 5 L into vacuum.​

Answer 1

Answer:

Explanation: Expansion of

a gas in vacuum (p  

ex

​

=0) is called free expansion

(−W)=p  

4

​

(V  

2

​

−V  

1

​

)=0×(5−1)=0

For isothermal

expansion q=0

By first law

of thermodynamics q=ΔU+(−W)

⇒0=ΔU+0

S0,ΔU=0

Answer 2

Answer:

The work done and the change in internal energy for an ideal gas expands isothermally from $1$L to $5$ L into the vacuum is zero·

Explanation:

Given that,

No Of moles of an ideal gas $=$ $1$ mol

The initial volume of the gas $V_i$ $=$ $1$ L

The final volume of the gas $V_f$ $=$ $5$ L

It is mentioned in the question that, the gas expands isothermally in a vacuum· Expansion of gas in a vacuum is called free expansion·

For an isothermal free expansion of an ideal gas, the work done is

  $-\ W\ =\ P_{(ex)} \ (V_f\ -\ V_i)$

During a free expansion of gas, it expands without any external pressure· So the external pressure acting on the gas is zero·

That is,

$P_{(ex)} \ =\ 0$

On substituting the values we get,

 ⇒  $-\ W\ =\ 0 \ (5\ -\ 1)$

 ⇒  $-\ W\ =\ 0 \ (4)$

 ⇒     $W\ =\ 0$

Since the external pressure for a free expansion of a gas is zero, the work done will always be zero·

To find out the change in the internal energy,

From the first law of thermodynamics, we know that

   $\Delta U\ =\ q\ +\ W$

Since the process is an expansion process, the work done will always have a negative value·

That is,

W  $=$ $-$ ve

For an isothermal expansion process,

q $=$ $W$

Since the work done is zero, the heat exchange will also be zero·

That is,

q $=$ $0$

Therefore,

 ⇒  $\Delta U\ =\ q\ +\ -W$

 ⇒  $\Delta U\ =\ 0\ +\ 0$

 ⇒  $\Delta U\ =\ 0$

Hence,

The change in internal energy for an ideal gas expands isothermally from $1$L to $5$ L into the vacuum is zero·