Whenever you introduce a liquid in an evacuated container (say you introduce 1.0

ID: 1006461 • Letter: W

Question

Whenever you introduce a liquid in an evacuated container (say you introduce 1.00 g), some of the liquid instantaneously vaporizes leaving you with a smaller volume of liquid. To illustrate this phenomenon, estimate (within +/-2.5 C) the temperature at which 0.25 g water sample is in the vapor phase (hence, 0.75 g remains in the liquid phase and 0.25 g transfers to the vapor phase). Assume the evacuated container has a volume of 7.8 L. To solve this problem, you need to consult the table of water vapor pressures at the end of this homework. Assume water vapor to behave ideally and assume the density of water liquid to be approximately 1 g/cm3.

Explanation / Answer

Volume occupied by 0.25g water=0.25g*1cm3/g=0.25 cm3=0.25*0.001L=0.00025 L

Using ideal gas equation,

PV=nRT

N=PV/RT

Volume occupied by gas=Vg=total volume, V-volume occupied by water,Vw=7.8L-0.00025L=7.8 approx ,negligible water volume

V/R=7.8L/0.0821L atm/K.mol=95.00 atm/K.mol

No of moles in 0.25 g water=0.25 g/18g/mol=0.014 moles

P/T=nR/V=0.014 moles/(95.0 atm/K.mol)=1.46*10^-4 atm/K

We need to compare the P/T ratio from the table,

Temperature in K=273+temp,(deg C)

Temperature (K)

Pressure (atm)

P/T

273

0.00603

2.21*10^-5

278

0.00861

3.1*10^-5

283

0.01211

4.28*10^-5

288

0.01683

5.84*10^-5

293

0.02308

7.88*10^-5

298

0.03126

1.05*10^-4

303

0.04187

1.38*10^-4

308

0.05550

1.80*10^-4

313

0.07279

318

0.09458

323

0.12172

328

0.15532

333

0.19655

338

0.24676

343

0.30750

348

0.38039

353

0.46724

358

0.57053

363

0.69179

368

0.83407

P/T is between 303 K and 308K .

Or , 303K-305K precisely

Or,30 deg C-34 deg C