Question

Two litres of an ideal gas at a pressure of 10 \[atm\] expands isothermally into vacuum until its total volume is 10 litres.
1.How much heat is absorbed and how much work is done in the expansion.
2.If the same expansion takes place against a constant external pressure of 1atm?
3.If the same expansion takes place to a final volume of 10 litres conducted reversibly?

Answer 1

Hint: Thermodynamics is the branch of science that deals with heat, work done, temperature and transfer of energy from one form to another. The first law of thermodynamics states that energy is conserved and can neither be created nor be destroyed. 

Formula used:
First law states that \[\Delta U = Q - W\].  Work done is \[W = - P\Delta V\] and work done for reversible reaction is  \[{W_{rev}} = - 2.303PV\log \dfrac{{{V_f}}}{{{V_i}}}\]

Complete step-by-step answer:We have already learnt the hint that the first law of thermodynamics highlights the law of conservation of energy. The first law states that energy can neither be created nor be destroyed but can be converted from one form to another form, it is always conserved. 
Let us know about this in detail. The relationship between heat and other forms of energy is determined under the branch of science. It explains that thermal energy is converted to other forms of energy and used for various purposes. It also deals with work, temperature and energy and the transfer of energy from one place to another.
\[\Delta U = Q - W\] 
Where \[\Delta U\] is the change in internal energy and \[Q\] is the added heat and \[W\] is the work done. 
Work is the process done by a system and it is not contained in a system. In thermodynamics, work done by a system is the energy transferred by the system to its surroundings. It is calculated by the formula \[W = - P\Delta V\] where \[P\] is the pressure and \[\Delta V\] is the change in volume. The work done for a reversible reaction is given by \[{W_{rev}} = - 2.303PV\log \dfrac{{{V_f}}}{{{V_i}}}\]
For the above question,
1. Work is done by the system during expansion
We know that \[W = - P\Delta V\]
It is given that gas is expanding in vacuum which means \[P = 0\]
Thus, \[W = 0\]
So no work is done. 
We know that the first law of thermodynamics states that \[\Delta U = Q - W\]
It is given that the system is isothermal (\[\Delta T = 0\]), that menas \[\Delta U = 0\]
So, \[Q = W = 0\]. Hence no heat is absorbed.
2. We know that, 
\[W = - P\Delta V\], where
\[
 P = 1atm \\
 \Delta V = 10.2 = 8L \\
 \]
So, \[W = - 1 \times 8 = - 8atmL\]
As, \[Q = - W\]
We get, \[Q = - ( - 8) = 8atmL\]
3. According to the question, work done is reversible. The expression used will be;
\[{W_{rev}} = - 2.303PV\log \dfrac{{{V_f}}}{{{V_i}}}\]
Given: \[
 P = 1atm \\
  {V_f} = 10L \\
  {V_i} = 8L \\
 \]
So,\[
  {W_{rev}} = - 2.303 \times 1 \times 10\log \dfrac{{10}}{2} \\
  {W_{rev}} = - 2.303 \times 1 \times 10\log 5 \\
  {W_{rev}} = - 16.1J \\
 \] 
As, \[
  {Q_{rev}} = - {W_{rev}} \\
 {Q_{rev}} = 16.1atmL \\
 \]

Note: An ideal gas refers to a gas whose molecules do not attract or repel each other and the only interaction between the molecules is the elastic collision due to impact of each other or the elastic collision of the molecules with the walls of the container in which the gas is stored.