Use the formula PV=nRT to find out. P refers to pressure, V to volume, n refers to the number of moles of particle in the space, R is always a constant and T is the temperature. so since we assume that n,R and T stay constant, the volume at 3 bar will be 1/3x so that PV remains at its initial balue: x.
A. An increase in pressure from 2 ATM to 3 ATM will result in a decrease in volume of gas. B. An increase in pressure from 3 ATM to 4 ATM will result in a decrease in volume of gas. C. A decrease in pressure from 4 ATM to 1 ATM will result in an increase in volume of gas. D. An increase in pressure from 1 ATM to 3 ATM will result in a decrease in volume of gas.
Assuming a fixed amount of an ideal gas kept at constant temperature, then the volume is reduced to a third of its former amount when the pressure is tripled. P V = n R T = constant = k P1 V1 = k = P2 V2 P2 = 3 P1 3 P1 V2 = P1 V1 V2 V1 / 3
The pressure is reduced to one third of the original pressure. The pressure will stay the same you are only changing the volume
The boiling point of water at 3 bar is 143.6 degrees Celsius.
The pressure of the balloon increases threefold when it is squeezed to one-third its volume with no change in temperature. This is because pressure is inversely proportional to volume according to Boyle's Law, which states that when volume decreases, pressure increases.
this pump is senterfugal and content from 3 stage . stage one is normal pressure and stage 2 is high pressure max pressure is 30 bar and volume is 3000 liter per minute at 10 bar )stage 1
Type your answer here... i have a tank with 5.18m^3 volume . iwant to know how many liters of air with 8 bar in there?
A. An increase in pressure from 2 ATM to 3 ATM will result in a decrease in volume of gas. B. An increase in pressure from 3 ATM to 4 ATM will result in a decrease in volume of gas. C. A decrease in pressure from 4 ATM to 1 ATM will result in an increase in volume of gas. D. An increase in pressure from 1 ATM to 3 ATM will result in a decrease in volume of gas.
The equation is P1V1=P2V2. (P1 is pressure before the change, P2 is the pressure after, V1 is the volume before the change, and V2 is the volume after it.) So to solve it, it would be the same change no matter how much the volume and pressure were to begin with. The values are P1= 1 atmosphere (the pressure of air at sea level) V1= 1 Liter which would mean P2=3 atmospheres 1*1=3(V2) 1/3 Liter= V2. So the volume would be one third of what it was before the pressure was tripled.
Air at 10 celsius and 1 bar(absolute) has density 1.25 kg/m3. By 6 bar do you mean gauge or absolute pressure? To apply a pressure factor you must use absolute, ie the pressure above a vacuum. Normal atmospheric conditions are 1 bar (abs). If you mean 6 bar(abs) the density becomes 6 x 1.25 = 7.5 kg/m3. On the other hand if you mean 6 bar (gauge) this is 7 bar (abs) and the density would be 7 x 1.25 =8.75 kg/m3. Either way, you just divide the weight of air in kg by the density to get the volume.
3 bar
2gallons
1/3 of the initial volume (Boyle-Mariotte law).
PV=nRT If n,R,T are constant than the only thing that can change is P 3*V has to be cancelled out 1/3 * 3 = 1 1/3P * 3V = nRT
Believe it or not, we can die from too little oxygen AND too much oxygen. This concept involves partial pressures. If we first consider using standard air (i.e. 20% Oxygen and 80% Nitrogen): (1 bar = 1 kilogram of pressure per square centimetre) At the surface - Total Pressure of air = 1 bar - Pressure of Oxygen = 0.2 bar - Pressure of Nitrogen = 0.8 bar At 10m - Total Pressure of air = 2 bar - Pressure of Oxygen = 0.4 bar - Pressure of Nitrogen = 1.6 bar At 20m - Total Pressure of air = 3 bar - Pressure of Oxygen = 0.6 bar - Pressure of Nitrogen = 2.4 bar At 30m - Total Pressure of air = 4 bar - Pressure of Oxygen = 0.8 bar - Pressure of Nitrogen = 3.2 bar At 40m - Total Pressure of air = 5 bar - Pressure of Oxygen = 1.0 bar - Pressure of Nitrogen = 4.0 bar At 50m - Total Pressure of air = 6 bar - Pressure of Oxygen = 1.2 bar - Pressure of Nitrogen = 4.8 bar At 60m - Total Pressure of air = 7 bar - Pressure of Oxygen = 1.4 bar - Pressure of Nitrogen = 5.6 bar At 70m - Total Pressure of air = 8 bar - Pressure of Oxygen = 1.6 bar - Pressure of Nitrogen = 6.4 bar It is at the partial pressure of 1.6 bar (which occurs at 70m where oxygen becomes toxic. That is why most dive organisations recommend 50m as a maximum for recreational diving. However, if pure oxygen is used: At surface - Pressure of Oxygen = 1 bar At 10m - Pressure of oxygen = 2 bar The oxygen has already become toxic! Hope that answers your question. The concept is called "partial pressures" and "oxygen toxicity" if you want to research more on a search engine.
From Boyle's law pressure (P) times volume (V) divided by temperature T is a constant; so if T is held constant then if pressure triples volume is decreased to 1/3 its original value
A quantitative answer cannot be calculated unless the units for pressure and volume are provided.