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# Solution for Derive an Expression for Maximum Work in Isothermal Reversible Expansion of Two Moles of an Ideal Gas. - HSC Science (Electronics) 12th Board Exam - Chemistry

ConceptChemical Thermodynamics and Energetic First Law of Thermodynamics

#### Question

Derive an expression for maximum work in isothermal reversible expansion of two moles of an ideal gas.

#### Solution

a. Consider ‘n’ moles of an ideal gas placed in a cylinder fitted with weightless, frictionless piston at a constant temperature ‘T’. If the gas expands isothermally and
reversibly from initial volume V1 to final volume V2 and if the expansion is by a small volume ‘dV’, then opposing pressure ‘P’ decreases by a small amount ‘dP’ producing small work done ‘dW’. Now, the pressure (pex) will be ‘P – dP’.
b.  This work done is given as:
Work done (dW) = pressure x change in volume
= -(P - dP) - dV
= -(PdV - dPdV)

both dP and dV are small, dP.dV is also very small and is neglected.
The above equation becomes dW =- PdV ….(1)
c.  When the expansion of a gas is carried out reversibly, there will be series of such ‘PdV’
terms. As the pressure decreases continuously, volume gradually increases from initial
volume ‘V1’ to final volume ‘V2’.
Thus, in order to obtain maximum work (Wmax), integrate equation (1) between the limits V1 to V2 as

int_0^(wmax)dw=-int_(v_1)^(v_2)Pdv ..........(2)

d. The general ideal gas equation for n mole is, PV = nRT

  or P=(nRT)/V

Substituting (3) in equation (2),

int_0^(wmax)dw=-int_(v_1)^(v_2)(nRT)/Vdv

Since the number of moles is constant, R is a gas constant and the process is isothermal,
therefore, T is constant. Hence, nRT = constant.

int_0^(wmax)dw=-nRTint_(v_1)^(v_2)1/Vdv

[w]_0^(wmax)=-nRTlog[V]_(v_1)^(v_2)

[W_(max)-0]=-nRT[log(V_2)-log(V_1)]

W_(max)=-nRTlog(V_2/V_1)

e. W_(max)=-2.303nRTlog(V_2/V_1)

f. Since, number of moles of ideal gas (n) = 2

W_(max)=-4.606xxRT log_10(V_2/V_1)`

Is there an error in this question or solution?

#### APPEARS IN

2014-2015 (October) (with solutions)
Question 2.5 | 2.00 marks
2016-2017 (March) (with solutions)
Question 4.2 | 7.00 marks

#### Video TutorialsVIEW ALL [2]

Solution Derive an Expression for Maximum Work in Isothermal Reversible Expansion of Two Moles of an Ideal Gas. Concept: Chemical Thermodynamics and Energetic - First Law of Thermodynamics.
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