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Question: Five moles of an ideal gas expands isothermally and reversibly at 300 K from initial volume\[{m^3}\]...

Five moles of an ideal gas expands isothermally and reversibly at 300 K from initial volumem3{m^3} 3 to a final volume by doing 20.92 kJ of work. Calculate the final volume. (R = 8.314 J/k mol.)

Explanation

Solution

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=−2.303nRTlogV1V2W = - 2.303nRTlog\dfrac{{{V_1}}}{{{V_2}}}

Complete step by step answer:
For an ideal gas, the equation is, PV=nRTPV = 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=−2.303nRTlogV1V2W = - 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=−2.303nRTlogV1V2 ⇒20.92×103=−2.303×5×8.314×300×log3V2 ⇒0.728=−log3V2 ⇒10−0.728=3V2 ⇒V2=310−0.728 ⇒V2=16.04  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 1604m31604{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=nRTPV = 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.