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Question: One mole of a gas is allowed to expand isothermally and reversibly from a volume of \[1d{m^3}\] to \...

One mole of a gas is allowed to expand isothermally and reversibly from a volume of 1dm31d{m^3} to 50dm350d{m^3} at 273K. Calculate W, ΔE\Delta E and q assuming the idea behaviour of the gas

Explanation

Solution

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=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=−2.303nRTlogV1V2....(1)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=−2.303RTlog150 W=−2.303×8.314×273log150 W=−5227.16×log150 W=5227.16×1.69 W=8.88 kJ/mol  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, ΔE\Delta E is equal to zero.
Now the 1st law of thermodynamics,
ΔE=q   +  w\Delta E = q\,\,\, + \,\,w , calculate the value of q as follows,

ΔE=q   +  w 0=q   +  w q= −w  q= −8.8 kJ/mol   \Delta E = q\,\,\, + \,\,w \\\ 0 = q\,\,\, + \,\,w \\\ q = \, - w\, \\\ q = \, - 8.8\,kJ/mol\, \\\

So, the value of W ΔE\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=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.