Question

Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until total volume is 10 litres. How much heat is absorbed and how much work is done in expansion?

Answer 1

2 litres of an ideal gas : V1 = 2 litres
P1 = 10 atms
T = constant
V2 = 10 litres
   =>  P2 = P1 V1 / V2 = 2 atm

work done =
$W=\int\limits^{V_2}_{V_1} {P} \, dV \\\\W=\int\limits^{V_2}_{V_1} {\frac{P_1*V_1}{V}} \, dV = 20\ [Ln\ V]_{V_1}^{V_2}\\\\=20*[Ln\ \frac{V_2}{V_1}]\ atm-litres\\\\=20*Ln_e\ 5\ *1.013*10^5\ Pa*10^{-3}m^3\\=3260.7 Joules$

This is the work done in expansion of the gas.  As the gas remains at the same temperature, there is no change in the internal energy.

So energy absorbed is = 3.2607 kJoules.

Answer 2

Answer

2 litres of an ideal gas : V1 = 2 litres

P1 = 10 atms

T = constant

V2 = 10 litres

=> P2 = P1 V1 / V2 = 2 atm

work done =

\begin{gathered}W=\int\limits^{V_2}_{V_1} {P} \, dV \\\\W=\int\limits^{V_2}_{V_1} {\frac{P_1*V_1}{V}} \, dV = 20\ [Ln\ V]_{V_1}^{V_2}\\\\=20*[Ln\ \frac{V_2}{V_1}]\ atm-litres\\\\=20*Ln_e\ 5\ *1.013*10^5\ Pa*10^{-3}m^3\\=3260.7 Joules\end{gathered}

W=

V

1

∫

V

2

PdV

W=

V

1

∫

V

2

V

P

1

∗V

1

dV=20 [Ln V]

V

1

V

2

=20∗[Ln

V

1

V

2

] atm−litres

=20∗Ln

e

5 ∗1.013∗10

5

Pa∗10

−3

m

3

=3260.7Joules

This is the work done in expansion of the gas. As the gas remains at the same temperature, there is no change in the internal energy.

So energy absorbed is = 3.2607 kJoules.