Question

find q for the evaporation of 385 g of water at this temperature. express your answer to three significant figures and include the appropriate units. q = 870 kj correct part b find w for the evaporation of 385 g of water at this temperature. assume that the density of the compound in its liquid state is 1.00 g ml⁻¹. express your answer to three significant figures and include the appropriate units. w = value units submit request answer part c find δu for the evaporation of 385 g of water at this temperature. assume that the density of the compound in its liquid state is 1.00 g ml⁻¹. express your answer to three significant figures and include the appropriate units. δu = - 64900 kj incorrect; try again; 2 attempts remaining

Explanation:

Step1: Calculate the volume of liquid water

The density formula is $
ho=\frac{m}{V}$, so $V_{liquid}=\frac{m}{
ho}$. Given $m = 385\ g$ and $
ho=1.00\ g/mL$, then $V_{liquid}=\frac{385\ g}{1.00\ g/mL}=385\ mL = 0.385\ L$.

Step2: Assume water vapor is an ideal gas

At standard - conditions for vaporization, we use the ideal gas law $PV = nRT$. For water, the molar mass $M = 18.02\ g/mol$, so the number of moles $n=\frac{m}{M}=\frac{385\ g}{18.02\ g/mol}\approx21.37\ mol$. At standard temperature and pressure for vaporization (assume $T = 373\ K$, $P = 1\ atm$), from $PV=nRT$ ($R = 0.0821\ L\cdot atm/(mol\cdot K)$), the volume of vapor $V_{vapor}=\frac{nRT}{P}=\frac{21.37\ mol\times0.0821\ L\cdot atm/(mol\cdot K)\times373\ K}{1\ atm}\approx652\ L$.

Step3: Calculate the work done

The work done in an isobaric process is $w=-P\Delta V$. $\Delta V=V_{vapor}-V_{liquid}\approx652\ L - 0.385\ L\approx652\ L$. Since $1\ L\cdot atm = 101.3\ J$, $w=-P\Delta V=- 1\ atm\times652\ L=-652\ L\cdot atm=-652\times101.3\ J\approx - 66.1\ kJ$.

Step4: Calculate the change in internal energy

The first - law of thermodynamics is $\Delta U=q + w$. We know from part A that $q = 870\ kJ$ and from step 3 that $w=-66.1\ kJ$. So $\Delta U=870\ kJ-66.1\ kJ = 804\ kJ$.

Answer:

For Part B:
$w=-66.1\ kJ$
For Part C:
$\Delta U = 804\ kJ$