The coefficient of thermal expansion, α, is nearly constant for a liquid α=(1/V)((dV/dT))P At 293 K and 1 atm for water α = 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?

Question

The coefficient of thermal expansion, α, is nearly constant for a liquid α=(1/V)((dV/dT))P At 293 K and 1 atm for water α = 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

Tardigrade Admin · Accepted Answer

dV=α VdT Δ W=-∫ PdV=-α V× P ∫T1T2 dT or Δ W=-2.1×18×10-6×105×10 J or Δ W=-2.1×18 J or Δ W=-37.8 J

$- 78.25 k J$

$78.15 J$

$- 37.8 J$

$- 83.14 J$

$d V = α V d T$
$Δ W = - \int P d V = - α V \times P \int_{T_{1}}^{T_{2}} d T$
or $Δ W = - 2.1 \times 18 \times 1 0^{- 6} \times 1 0^{5} \times 10 J$
or $Δ W = - 2.1 \times 18 J$
or $Δ W = - 37.8 J$

Q. The coefficient of thermal expansion, α, is nearly constant for a liquid α=V1​(dTdV​)P​ At 293K and 1atm for water α=2.1K−1 What is the approximate work done when 1 mole of liquid is heated from 288K to 298K at 1atm ? Molar volume of liquid =18ml?

−78.25kJ

78.15J

−37.8J

−83.14J

dV=αVdT ΔW=−∫PdV=−αV×P∫T1​T2​​dT or ΔW=−2.1×18×10−6×105×10J or ΔW=−2.1×18J or ΔW=−37.8J

$- 78.25 k J$

$78.15 J$

$- 37.8 J$

$- 83.14 J$

$d V = α V d T$
$Δ W = - \int P d V = - α V \times P \int_{T_{1}}^{T_{2}} d T$
or $Δ W = - 2.1 \times 18 \times 1 0^{- 6} \times 1 0^{5} \times 10 J$
or $Δ W = - 2.1 \times 18 J$
or $Δ W = - 37.8 J$