Question

The coefficient of thermal expansion, a, is nearly constant for a liquid a = $\left( \frac{dV}{dT} \right)_{P}$

At 293 K and 1 atm for water a = 2.1 K^–1

What is the approximate work done when 1 mole of liquid is heated from 288 K to 298 K at 1 atm? Molar volume of liquid = 18 ml

• A. –78.25 kJ
• B. 78.15 J
• C. –37.8 J
• D. –83.14 J

Answer 1

Answer: (C)

–37.8 J

Answer 2

dV = a VdT

DW = –$\int_{}^{}{PdV = –\alpha}V \times P\int_{T_{1}}^{T_{2}}{dT}$

or DW = – 2.1 × 18 × 10^–6 × 10⁵ × 10 J

or DW = – 2.1 × 18 J

or DW = – 37.8 J