Answer:
- final temperature (T2) = 748.66 K
- ΔU = w = 5620.26 J
- ΔH = 9367.047 J
- q = 0
Explanation:
ideal gas:
reversible adiabatic compression:
∴ q = 0
∴ w = - PδV
⇒ δU = δw
⇒ CvδT = - PδV
ideal gas:
⇒ PδV + VδP = RδT
⇒ PδV = RδT - VδP = - CvδT
⇒ RδT - RTn/PδP = - CvδT
⇒ (R + Cv,m)∫δT/T = R∫δP/P
⇒ [(R + Cv,m)/R] Ln (T2/T1) = Ln (P2/P1) = Ln (1 E6/1 E5) = 2.303
∴ (R + Cv,m)/R = (R + (3/2)R)/R = 5/2R/R = 2.5
⇒ Ln(T2/T1) = 2.303 / 2.5 = 0.9212
⇒ T2/T1 = 2.512
∴ T1 = 298 K
⇒ T2 = (298 K)×(2.512)
⇒ T2 = 748.66 K
⇒ ΔU = Cv,mΔT
⇒ ΔU = (3/2)R(748.66 - 298)
∴ R = 8.314 J/K.mol
⇒ ΔU = 5620.26 J
⇒ w = 5620.26 J
⇒ ΔH = ΔU + nRΔT
⇒ ΔH = 5620.26 J + (1 mol)(8.314 J/K.mol)(450.66 K)
⇒ ΔH = 5620.26 J + 3746.787 J
⇒ ΔH = 9367.047 J
hi im breanna
Answer:
The mole is simply a very large number that is used by chemists as a unit of measurement.
Explanation:
The mole is simply a very large number,
6.022
×
10
23
, that has a special property. If I have
6.022
×
10
23
hydrogen atoms, I have a mass of 1 gram of hydrogen atoms . If I have
6.022
×
10
23
H
2
molecules, I have a mass of 2 gram of hydrogen molecules. If I have
6.022
×
10
23
C
atoms, I have (approximately!) 12 grams.
The mole is thus the link between the micro world of atoms and molecules, and the macro world of grams and litres, the which we can easily measure by mass or volume. The masses for a mole of each element are given on the periodic table as the atomic weight. So, if have 12 g of
C
, I know, fairly precisely, how many atoms of carbon I have. Given this quantity, I know how many molecules of
O
2
are required to react with the
C
, which I could measure by mass or by volume.
The atomic mass would not change since the mass of an electron is negligible compared to the mass of protons and neutrons
