What compound consists of 0.248 g magnesium, 0.655 g sulfur, and 1.31 g oxygen
1 answer:
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Answer:
A. The pressure will increase 4 times. P₂ = 4 P₁
B. The pressure will decrease to half its value. P₂ = 0.5 P₁
C. The pressure will decrease to half its value. P₂ = 0.5 P₁
Explanation:
Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.
<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>
<em />
<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>
V₂ = 0.25 V₁. According to Boyle's law,
P₁ . V₁ = P₂ . V₂
P₁ . V₁ = P₂ . 0.25 V₁
P₁ = P₂ . 0.25
P₂ = 4 P₁
<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>
T₂ = 0.5 T₁. According to Gay-Lussac's law,
<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>
n₂ = 0.5 n₁.
P₁ in terms of the ideal gas equation is:
P₂ in terms of the ideal gas equation is:
Answer:
we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:
ln (P2/P1) = -
)
where
P1 and P2 are the vapour pressures at temperatures T1 and T2
Δ vapH = the enthalpy of vaporization of the ETHANOL
R = the Universal Gas Constant
In this problem,
P 1 = 100 mmHg
; T 1 = 34.7 °C = 307.07 K
P 2 = 760mmHg
T 2 =T⁻²=?
Δ vap H = 38.6 kJ/mol
R = 0.008314 kJ⋅K -1 mol -1
ln ( 760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1 /0.008314 )
0.0004368=(0.00325 - T⁻²)
T⁻²=0.002813
T² = 355.47K