The new pressure changed relative to the original by a factor of 8/3.
<h3>What is the change in the pressure?</h3>
Using the ideal gas equation;
PV = nRT
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
P1 = nRT/V
P2 = 4n * R * 2T/3V
Hence;
P2/P1 = 4n * R * 2T/3V ÷ nRT/V
P2/P1 = 4n * R * 2T/3V * V/nRT
P2/P1 =4 * 2/3
= 8/3
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Answer:
1 mole of potassium hydroxide, KOH
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
KOH + HCl —> KCl + H₂O
Considering the balanced equation above, we can see clearly that 1 mole of potassium hydroxide, KOH reacted with 1 mole of hydrochloric acid, HCl.
Therefore, we can conclude that at every given point, 1 mole of potassium hydroxide, KOH is required to react with 1 mole of hydrochloric acid, HCl.
Answer:
What language is this please i need help
1 answer · Chemistry
Best Answer
Water steam condenses if its pressure is equal to vapor saturation vapor pressure.
Use the Clausius-Clapeyron relation.
I states the temperature gradient of the saturation pressure is equal to the quotient of molar enthalpy of phase change divided by molar volume change due to phase transition time temperature:
dp/dT = ΔH / (T·ΔV)
Because liquid volume is small compared to vapor volume
ΔV in vaporization is approximately equal to to the vapor volume. Further assume ideal gas phase:
ΔV ≈ V_v = R·T/p
Hence
dp/dT = ΔHv / (R·T²/p)
<=>
dlnp/dT = ΔHv / (R·T²)
If you solve this DE an apply boundary condition p(T₀)= p₀.
you get the common form:
ln(p/p₀) = (ΔHv/R)·(1/T₀ - 1/T)
<=>
p = p₀·exp{(ΔHv/R)·(1/T₀ - 1/T)}
For this problem use normal boiling point of water as reference point:
T₀ =100°C = 373.15K and p₀ = 1atm
Therefore the saturation vapor pressure at
T = 350°C = 623.15K
is
p = 1atm ·exp{(40700J / 8.314472kJ/mol)·(1/373.15K - 1/623.15K)} = 193 atm
hope this helps
