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:
you can see the answer at the pic
Answer:
Due to how big the glacier is you would need more heat to melt it and since the ice cube is smaller a small amount of heat would melt it
Explanation:
Answer:
<em>Dry suction chest tube system</em>
Explanation:
<em>The dry suction drains that are self-regulating today use a small, adjustable regulator installed into the drain. </em>
So long as there is enough air flow from the suction of the wall (that will be shown on the drain) it will respond correctly to changes in the pressure of the source or the patient to retain suction at the drain level.
It's also a quiet mode of activity.
Answer:
Option A = 14.19 atm
Explanation:
Given data:
Volume of gas = 2.50 L
Number of moles of gas = 1.35 mol
Temperature of gas = 320 K
Pressure of gas = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will put the values in formula.
P × 2.50 L = 1.35 mol × 0.0821 atm.L/ mol.K× 320
P = 35.467 atm.L/ 2.50 L
P = 14.19 atm
Thus, option A is correct.