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← IIT-JEE 2010 · Paper 2
ChemistryInteger+3Question 11 of 57

IIT-JEE 2010 · Paper 2, Chemistry question 11

One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is wsw_s and that along the dotted line path is wdw_d, then the integer closest to the ratio wd/wsw_d/w_s is

P (atm) against V (lit). Solid path: a smooth isotherm-like curve from a (0.5, 4.0) through (1.0, 2.0), (2.0, 1.0), (4.0, 0.5) to b (5.5, ≈0.36). Dashed path: from a horizontally to (2.0, 4.0), down to (2.0, 1.0), across to (3.0, 1.0), down to (3.0, ≈0.67), then across to (5.5, ≈0.67) above b.
Answer and worked solution

Official answer: 2

Solid path. The points (0.5, 4), (1, 2), (2, 1), (4, 0.5) all have PV=2PV=2 L atm: an isotherm, ending at V=5.5V=5.5 L:
ws=PVln⁡VbVa=2ln⁡5.50.5=2ln⁡11≈4.80 L atm.w_s=PV\ln\frac{V_b}{V_a}=2\ln\frac{5.5}{0.5}=2\ln11\approx4.80\text{ L atm}.

Dashed path. Work is done only on the horizontal (isobaric) steps:
wd=4.0(2.0−0.5)+1.0(3.0−2.0)+23(5.5−3.0)=6+1+1.67≈8.67 L atm.w_d=4.0(2.0-0.5)+1.0(3.0-2.0)+\tfrac23(5.5-3.0)=6+1+1.67\approx8.67\text{ L atm}.

Ratio. wdws≈1.81\dfrac{w_d}{w_s}\approx1.81, closest integer 2.

Answer: 2

Topics:First law and enthalpy
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