One mole of a gas is allowed to expand isothermally and reversibly from a volume of \[1d{m^3}\] to \[50d{m^3}\] at 273K. Calculate W, \[\Delta E\] and q assuming the idea behaviour of the gas
Answer
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Hint: 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.
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,
\[
W = - 2.303RTlog\dfrac{1}{{50}} \\
W = - 2.303 \times 8.314 \times 273log\dfrac{1}{{50}} \\
W = - 5227.16 \times log\dfrac{1}{{50}} \\
W = 5227.16 \times 1.69 \\
W = 8.88\,kJ/mol \\
\]
Now, for an ideal gas, the change of internal energy depends upon the temperature only. Therefore, at the isothermal process, \[\Delta E\] is equal to zero.
Now the 1st law of thermodynamics,
\[\Delta E = q\,\,\, + \,\,w\] , calculate the value of q as follows,
\[
\Delta E = q\,\,\, + \,\,w \\
0 = q\,\,\, + \,\,w \\
q = \, - w\, \\
q = \, - 8.8\,kJ/mol\, \\
\]
So, the value of W \[\Delta E\] and q are 8.8kJ/mol. 0kJ/mol and -8.8kJ/mol respectively.
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.
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,
\[
W = - 2.303RTlog\dfrac{1}{{50}} \\
W = - 2.303 \times 8.314 \times 273log\dfrac{1}{{50}} \\
W = - 5227.16 \times log\dfrac{1}{{50}} \\
W = 5227.16 \times 1.69 \\
W = 8.88\,kJ/mol \\
\]
Now, for an ideal gas, the change of internal energy depends upon the temperature only. Therefore, at the isothermal process, \[\Delta E\] is equal to zero.
Now the 1st law of thermodynamics,
\[\Delta E = q\,\,\, + \,\,w\] , calculate the value of q as follows,
\[
\Delta E = q\,\,\, + \,\,w \\
0 = q\,\,\, + \,\,w \\
q = \, - w\, \\
q = \, - 8.8\,kJ/mol\, \\
\]
So, the value of W \[\Delta E\] and q are 8.8kJ/mol. 0kJ/mol and -8.8kJ/mol respectively.
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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