Five moles of an ideal gas expands isothermally and reversibly at 300 K from initial volume\[{m^3}\] 3 to a final volume by doing 20.92 kJ of work. Calculate the final volume. (R = 8.314 J/k mol.)
Answer
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Hint: Depending on the value of the variables the thermodynamic process can be of various types, like, isothermal, isobaric, isochoric, adiabatic, etc. For an ideal gas, from the ideal gas law, P-V remains constant through an isothermal process. For an isothermal, reversible process, the work done by the gas is equal to the area under the relevant pressure-volume isotherm.
Formula used: \[W = - 2.303nRTlog\dfrac{{{V_1}}}{{{V_2}}}\]
Complete step by step answer:
For an ideal gas, the equation is, \[PV = nRT\]. Where p is the pressure, v is the volume, R is the gas constant and T is the temperature n is the number of moles.
The work is done formula isothermal reversible process is,
\[W = - 2.303nRTlog\dfrac{{{V_1}}}{{{V_2}}}\] …. (1)
Where, w is the work done.
Now from the given values calculate the work done is,
\[
W = - 2.303nRTlog\dfrac{{{V_1}}}{{{V_2}}} \\
\Rightarrow 20.92 \times {10^3} = - 2.303 \times 5 \times 8.314 \times 300 \times log\dfrac{3}{{{V_2}}} \\
\Rightarrow 0.728 = - log\dfrac{3}{{{V_2}}} \\
\Rightarrow {10^{ - 0.728}} = \dfrac{3}{{{V_2}}} \\
\Rightarrow {V_2} = \dfrac{3}{{{{10}^{ - 0.728}}}} \\
\Rightarrow {V_2} = 16.04 \\
\]
So, the final volume is \[1604{m^3}\].
Note: Van Der Waals equation of state for real gases is obtained from the modification of ideal gas equation or law. According to an ideal gas equation, \[PV = nRT\] where P is the pressure, V is the volume, T is the temperature, R is the universal gas constant and n is the number of moles of an ideal gas.
The Vander Waals equation considers the molecular interaction forces i.e. both the attractive and repulsive forces and the molecular size. There comes a volume correction in the Vander Waal equation. As the particles have a definite volume, the volume available for the movement of molecules is not the entire volume of the container but less than that.
Formula used: \[W = - 2.303nRTlog\dfrac{{{V_1}}}{{{V_2}}}\]
Complete step by step answer:
For an ideal gas, the equation is, \[PV = nRT\]. Where p is the pressure, v is the volume, R is the gas constant and T is the temperature n is the number of moles.
The work is done formula isothermal reversible process is,
\[W = - 2.303nRTlog\dfrac{{{V_1}}}{{{V_2}}}\] …. (1)
Where, w is the work done.
Now from the given values calculate the work done is,
\[
W = - 2.303nRTlog\dfrac{{{V_1}}}{{{V_2}}} \\
\Rightarrow 20.92 \times {10^3} = - 2.303 \times 5 \times 8.314 \times 300 \times log\dfrac{3}{{{V_2}}} \\
\Rightarrow 0.728 = - log\dfrac{3}{{{V_2}}} \\
\Rightarrow {10^{ - 0.728}} = \dfrac{3}{{{V_2}}} \\
\Rightarrow {V_2} = \dfrac{3}{{{{10}^{ - 0.728}}}} \\
\Rightarrow {V_2} = 16.04 \\
\]
So, the final volume is \[1604{m^3}\].
Note: Van Der Waals equation of state for real gases is obtained from the modification of ideal gas equation or law. According to an ideal gas equation, \[PV = nRT\] where P is the pressure, V is the volume, T is the temperature, R is the universal gas constant and n is the number of moles of an ideal gas.
The Vander Waals equation considers the molecular interaction forces i.e. both the attractive and repulsive forces and the molecular size. There comes a volume correction in the Vander Waal equation. As the particles have a definite volume, the volume available for the movement of molecules is not the entire volume of the container but less than that.
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