Question

The volume of 1g of liquid mercury is 0.07355 mL at 0 C and 0.07489 at 100 C. The molar heat capacity of mercury over this range is 6.60 cal deg-1 mole-1. Use these data to calculate w=P(delta V), q, and delta E for the process.

Hg(0 C, 1 atm) -> Hg (100 C, 1 atm)

Answers from back of book: w= 6.50*10^-3 cal , q = 660 cal, delta E= 660 cal

I cannot figure out how to start these problems. Thank you.

Answer 1

Hg(0℃, 1atm, 1mol) Hg(100℃, 1atm, 1mol)

Volume of 1g of Hg at 0℃ = 0.07355ml = 7.355×10^-5L

Volume of 1g of Hg at 100℃ = 0.07489ml = 7.489×10^-5L

Volume of 1 mole of Hg = volume of 1g × molar mass of Hg

Molar mass of Hg= 200.59g/mol

Volume of 1 mol Hg, V₂ = 7.489×10^-5× 200.59 L

Volume of 1mol Hg, V₁= 7.355×10^-5×200.59 L

Work done = - P_ext(V₂-V₁)

= - 1atm ×( 7.489×10^-5 - 7.355×10^-5) × 200.59 L

= - 26.879×10^-5atm-L

1atm-L = 101325N/m² × 10^-3 m³ = 101.325 N.m = 101.325J

Work done = - 26.879×10^-5 × 101.325 J = 2.72352×10^-2 J

Work done = -( 2.72352×10^-2 / 4.184)cal = - 6.50×10^-3 Cal.

Negative sign just indicate that expansion work done take place.

Work done = 6.50×10^-3 Cal. (answer)

Heat = q = number of moles× molar heat capacity× ∆T

= 1mol × 6.60cal/℃.mol × (100-0)℃ = 660 cal . (answer)

According to first law of thermodynamics : ∆E = q + w

∆E = 660 cal + (-6.50×10^-3cal) = 659.9935 cal = 660 cal. (answer)