Answer:

General Formulas and Concepts:
<u>Chemistry - Gas Laws</u>
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
- Charles' Law:

Explanation:
<u>Step 1: Define</u>
Initial Volume: 5.0 L H₂ gas
Initial Temp: 273 K
Final Temp: 985 K
Final Volume: ?
<u>Step 2: Solve for new volume</u>
- Substitute:

- Cross-multiply:

- Multiply:

- Isolate <em>x</em>:

- Rewrite:

<u>Step 3: Check</u>
<em>We are given 2 sig figs as the smallest. Follow sig fig rules and round.</em>
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Answer:
6,461.44 sana po makatulong
This problem is to use the Claussius-Clapeyron Equation, which is:
ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]
Where p2 and p1 and vapor pressure at estates 2 and 1
ΔH is the enthalpy of vaporization
R is the universal constant of gases = 8.314 J / mol*K
T2 and T1 are the temperatures at the estates 2 and 1.
The normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa
Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol
=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]
=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x = 0.000157 + 1/330.95 = 0.003179
=> x = 314.6 K => 314.6 - 273.15 = 41.5°C
Answer: 41.5 °C
We can use the ideal gas equation:
PV = nRT
P = 202.6kPa = 202600 Pa (You have to
multiply by 1000)
n = 0.050 mole
R = 0.082 atm*l/(K*mol)
T = 400K
We will have to convert from Pa to atm or
viceversa.
101325 Pa________1 atm
202600 Pa________x = 2.00 atm
2atm*V = 0.050 mole*0.082 atm*l/(K*mol)* 400K
V = 0.050 mole*0.082 atm*l/(K*mol)* 400K/2atm
= 0.82 liters = 820 mililiters