Question

2L of an ideal gas at a pressure of 10 atm expansion isothermally into a vacuum until its total volume is 1OL
(i) How much heat is absorbed and how much work is done in the expansion?
(ii) How much heat is absorbed if this system expands against a constant external pressure of 1 atm?
(iii) How much heat is absorbed if the expansion is conducted reversibly at 298K.

Answer 1

Hint: An isothermal process is a change in the system such that the temperature to the system remains constant. In other words, in an isothermal process, \[{{\Delta T = 0}}{{.}}\] Since \[{{\Delta T = 0}}{{.}}\], the change in internal energy is also zero. In a pressure- volume isotherm, the work done by the gas is equal to the area under the relevant pressure -volume isotherm

Complete step by step answer:
(i) work done during an expansion, we know that isothermal work done,
\[{{W = - }}{{{P}}_{{{ext}}}}{{.}}{{\Delta V}}\]
The given gas is expanding in vacuum their fore, the pressure is zero i.e.,
\[{{{P}}_{{{ext}}}}{{ = }}\,\,{{0}}\], substituting in the above equation,
\[{{W = 0}}\], This means that no work is done.
From first law of thermodynamics we have,
\[{{\Delta U = q + W}}\], as the system is working at constant temperature, there is no change in internal energy,
\[{{\Delta U = 0}}\], Hence \[{{q}}\,\,{{ = }}\,\, - {{W}}\,\,{{ = }}\,\,{{0}}\]
So, no heat is absorbed during the expansion.
(ii) Isothermal work done,\[{{W = - }}{{{P}}_{{{ext}}}}{{.}}{{\Delta V}}\]
\[{{{P}}_{{{ext}}}}{{ = 1atm}}\] and the change in volume is, \[{{\Delta V}}\,\,{{ = }}\,\,{{10}}\,\,{{ - }}\,\,{{2L}}\,\,{{ = }}\,\,{{8L}}\]
\[{{W = - }}{{{P}}_{{{ext}}}}{{.}}{{\Delta V}}\]\[{{ = }}\,\,{{ - 1 \times 8}}\,\,{{ = }}\,\,{{ - 8J}}\]
As, The heat absorbed, \[{{q}}\,\,{{ = }}\,\, - {{W}}\,\], \[{{q}}\,\,{{ = }}\,{{ - }}\left( {{{ - 8J}}} \right)\]\[{{ = }}\,8\]
(iii) the work is done reversibly, work done is given by the equation,
\[{{{W}}_{{{rev}}}}{{ = - 2}}{{.303nRTlog}}\left( {\dfrac{{{{{V}}_{{f}}}}}{{{{{V}}_{{i}}}}}} \right)\]
\[{{{W}}_{{{rev}}}}{{ = - 2}}{{.303PVlog}}\left( {\dfrac{{{{10}}}}{{{2}}}} \right)\]​
\[{{{W}}_{{{rev}}}}\,\,{{ = }}\,\,{{ - 2}}{{.303 \times 1 \times 10 \times log5}}\,\,{{ = }}\,\,\,{{ - 16}}{{.1Latm}}\] [Given P=1 atm in (ii) part]
As \[{{{q}}_{{{rev}}}}\,\,{{ = }}\,\,{{ - }}{{{W}}_{{{rev}}}}\,\]\[{{ = }}\,\,{{16}}{{.1}}\,\,{{L - atm}}\]

Note: The work done during the free expansion of an ideal gas is zero in both reversible and irreversible process. In an isothermal process, the cylinder must have conducting walls
In a thermodynamic process, adiabatic expansion for an ideal gas in a closed system, is the condition, pressure is constant and the temperature is decreasing. This change causes a change in temperature.