Entropy Change During Expansion Into Vacuum
Task number: 3940
Determine the entropy change of ideal gas with temperature of 20 °C, pressure of 100 kPa and volume of 2 l, provided that the gas expands into vacuum to twice its original volume. Consider the process to be isothermal.
Hint 1 – What process is it?
The process of gas expanding into vacuum is called a spontaneous expansion.
A spontaneous expansion is an irreversible adiabatic process during which work done on and by the gas is zero. Therefore it applies: Q = W = 0. From the First Law of Thermodynamics, the following applies for the internal energy of gas: ΔU = 0. Since the internal energy of gas depends only on its temperature and not on its volume, its temperature remains constant during a spontaneous expansion.
During a spontaneous expansion, the pressure and volume of the gas changes unpredictably. Only the initial and final states are in equilibrium.
Hint 2 – Entropy
Entropy S is a state function. What does it mean?
Hint 3 – Entropy Change
Entropy change ΔS of a system during a process starting in equilibrium state A and ending in equilibrium state B is described as
\[\mathrm{\Delta} S=S_B-S_A=\int\limits_{A}^{B}\frac{dQ}{T},\]where Q is heat tranferred in or out of the system during the process, and T is thermodynamic temperature of the system.
During spontaneous expansion, however, pressure, temperature and volume of the gas changes unpredictably. Therefore, we cannot find a relation between heat Q and temperature T which would enable us to perform given integration. How do we find the unknown entropy change then?
Hint 4 – Entropy Change of Reversible Isothermal Expansion
Entropy change ΔS of reversible isothermal expansion is described as follows:
\[\Delta S=\frac{Q}{T},\]where Q is total heat and T is thermodynamic temperature during this process.
Given Values
t = 20 °C => T = 293.15 K gas temperature p1 = 100 kPa = 105 Pa gas pressure V1 = 2 l = 2·10−3 m3 gas initial volume V2 = 2V1 gas final volume ΔS = ? entropy change Analysis
When gas expands into a vacuum, this process is called a spontaneous expansion. It is an irreversible adiabatic process during which the gas does not perform work and no work is supplied to the system. Furthermore, pressure and volume of the gas change unpredictably. Therefore, we cannot use the same formulas as with irreversible processes during calculation.
We use the fact that entropy is a state function. Its change between the initial and final state therefore depends only on these two states and not on the way the system got there. We substitute the irreversible spontaneous expansion with reversible isothermal expansion with the same initial and final state. This process is convenient because temperature does not change during spontaneous expansion of ideal gas.
Then we determine the entropy change for this chosen process as a ratio of total exchanged heat and thermodynamic temperature during this process. The heat supplied during reversible isothermal expansion is equal to work done by the gas. To determine this work we need to use integration due to the fact that pressure is a volume function. Pressure as a volume function is expressed from Boyle's Law.
Since entropy is a state variable, the determined entropy change is the same as the entropy change during spontaneous expansion.
Solution
Change in entropy ΔS of a system in a process starting in an equilibrium state A and ending in an equilibrium state is described as:
\[\mathrm{\Delta} S=S_B-S_A=\int\limits_{A}^{B}\frac{dQ}{T},\]where Q is heat transferred to or from the system during the process and T is thermodynamic temperature.
We need to determine entropy change during a spontaneous epxansion. During this process, however, pressure, temperature and volume changes unpredictably. Only the initial state A and final state B are at equilibrium. Therefore, we cannot find a relation between heat Q and temperature T that would enable us to perform the integration in the above mentioned formula.
However, we can use the fact that entropy is a state function. This means that its change between initial and final state depends only on these two states and not on the way the system got from one state to the other.
Let us replace the irreversible spontaneous expansion with a convenient reversible process. Since temperature does not change during a spontaneous expansion of ideal gas, we choose a reversible isothermal expansion with the same initial state A and final state B.
For entropy change during this process, the following simplified relation applies
\[\mathrm{\Delta} S=\frac{Q}{T}.\]Now we determine it.
During a reversible isothermal process, the internal energy of gas does not change which according to the First Law of Thermodynamics means that accepted heat Q is equal to work W done by the system during expansion.
This work can be determined by a relation
\[W=\int\limits_{V_1}^{V_2}p\,\text{d}V,\]where V1 and V2 is the initial and final volume of gas and p its pressure that changes during the expansion (it is a volume function).
We determine the pressure as a volume function from Boyle's Law:
\[p_1V_1=pV.\]From here we can determine pressure p:
\[p = \frac{p_1V_1}{V}.\]Now we can perform the integration:
\[W = \int\limits_{V_1}^{V_2}p\, \text{d}V = \int\limits_{V_1}^{2V_1}\frac{p_1V_1}{V}\, \text{d}V =\]we factor the constant values out of the integral
\[=p_1V_1 \int\limits_{V_1}^{2V_1}\frac{1}{V}\, \text{d}V = \]then we perform the integration an substitute the limits
\[=p_1V_1[\ln V]_{V_1}^{2V_1} = p_1V_1 \ln \frac{2V_1}{V_1}= p_1V_1 \ln 2.\]The entropy change we are looking for is:
\[\mathrm{\Delta} S=\frac{Q}{T}=\frac{W}{T} = \frac{p_1V_1\ln{2}}{T}.\]As we have already stated above, entropy is a state function. Therefore this determined entropy change is the same as entropy change during a spontaneous expansion.
Numerical Solution
\[\Delta S= \frac{p_1V_1\ln{2}}{T}= \frac{10^5\cdot{ 2}\cdot{ 10^{-3}}\cdot \ln{2}}{293.15} \,\mathrm{JK^{-1}}\dot{=} 0.47\,\mathrm{JK^{-1}}\]Answer
Entropy increased by approximately 0.47 JK−1 during the expansion.
