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:

<span>2 C6H6 + 15 O2 = 12 CO2 + 6 H2O
</span>
2 moles C6H6 -------------------- 6 moles H2O
8.73 moles C6H6 --------------- ( moles H2O)
( moles H2O) = 8.73 x 6 / 2
( moles H2O) = 52.38 / 2
= 26.19 moles of H2O
Molar mass H2O = 18.00 g/mol
1 mole -------------- 18.00 g
26.19 moles ----- ( mass of H2O)
mass of H2O = 26. 19 x 18.00 => 471.42 g of water
hope this helps!
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