What gas law applies to a can being crush from hot temperature to cold temperature?
157020
It is Boyle's law & Charle's Law combined\nVolume of a gas is inversly propotional to Temprature & Pressure.\nOn Cooling can the Pressure & volume is decreased on can\nSince the outer layer of can feels pressure of air & it is more than inner pressure of can So it is crushed
979941
Completely exothermic. The atoms are losing energy to get to a solid state. Gases have most energy since they are free and unbonded to neighbouring molecules. Liquids have enough energy to stay together but still move about one another. In solids they have just enough energy just to vibrate about in a fixed point, closely bonded to neighbouring atoms. As you can there is an energy loss as you go from gas state to solid state. Loss of energy is exothermic.\n\nYes it is a reaction to change of heat.
544724
are you talking about the ideal gas law? if so...\n\nThe ideal gas law is the equation of state of an ideal gas. The state of an amount of gas is determined by its pressure, volume, and temperature. The equation has the form\n\n    pV = nRT\n\nwhere\n\n    p:is the pressure (kPa),\n    V:is the volume (L),\n    n:is the number of moles of gas (mol),\n    R:is the gas constant, and\n    T:is the temperature (K).\n\nR is commonly approximated to 8.31 J·K-1·mol-1 when using Pa for pressure. However this can vary if using a different unit for pressure (see gas constant).\n\nso...\nV=nRT/p\nn=pV/RT\n\n\n\nor do you need density?\n\nDensity=Mass/Volume\n\n\n\nor\n\nVolume=length*width*height
1153522
The quantities you have listed are in fact related via the ideal gas law.  So, assuming you aren't in huge temperature or pressure extremes, you can use the ideal gas law with decent results.\n\nIdeal Gas Law\nPV = nRT\n\nwhere P = pressure (N/m^2)\nV = volume (m^3)\nn = number of moles of gas\nR = universal gas constant = 8.3145 J/(mol*K)\nT = temp (in Kelvins)\n\nSo, by looking at the equation you can not convert air pressure directly to volume without also knowing the temperature and number of moles of gas.
980849
The relationship between volume and pressure. Remember that theBoyles' law assumes the temperature to be constant.\n\n  P1V1 = P2V2\n\nV1 = original volume\n\nV2 = new volume\n\nP1 = original pressure\n\nP2 = new pressure
481468
Boyle's law states that the volume of a fixed mass of gas varies inversely with the pressure at constant temperature.\n\nUsing P1 and V1 to stand for initial conditions and P2 and V2 to stand for new conditions results in the following equation.\n\nP1xV1=P2xV2\n\nSimple proportion...\n\nNow, the law relating to pressure and temperature would be Gay-Lussac's law which states the pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature.\n\nFor a given mass of gas at constant volume, the ratio P/T is the same for any set of pressure-temperature values.\n\nThe proportion for Guy-Lussac's law would be:\n\nP1/T1=P2/T2\n\nSo, to answer your question, for Guy-Lussac's law, not Boyle's law, a reduction of pressure WOULD cause a reduction in temperature.
484846
1. The freezing point of water and gasoline can be compared in terms of their intermolecular forces. The stronger the intermolecular forces of a substance, the higher the freezing point (as more energy is required to overcome those forces). Water has a low molecular weight and so has weak dispersion forces. Gasoline consists of long chained hydrocarbons with high molecular weights and so would have much stronger dispersion forces than water. Even though water has strong hydrogen bonding, the overall intermolecular forces in gasoline would be much stronger and so it would have a lower freezing point, that is, less than 0 degrees celsius. \n\n2. Freezing and melting are opposite processes. So if water freezes at 0 degrees celsius, then it must melt at 0 degrees celsius.
278691
Use the combined law of gases formula P1*v1/T1=P2*v2/T2.  Where P=pressure, v=volume, and T=temperature.  If the problem doens't have either the temperatures or the pressures assume them to be 1.  I.E. The answer to problem number 7 is:  8 liters.
838291
Good question!\n\nFirst law of thermodynamics states that the internal energy change ΔE of a system is equal to (Q - W). Where Q is the heat absorbed by the system and W is the work done BY the system on the surroundings.\n\nNow, as you already might be knowing that work can also be expressed in terms of pressure and volume as-\n\nW = PΔV (where P is pressure of the gas inside the system and ΔV is the change of volume)\n\n[Here: "system" = balloon; and "surroundings" = refrigerator]\n\nAnd as the change in volume (ΔV) in the case of balloon stated above is negative i.e it deflates, work must be negative i.e (minus W).\n\nNow, as Q is also negative (heat is given off by the balloon), the equation of First Law of thermodynamics reads-\n\nΔE = -Q - (-W)\nΔE = W - Q\n\nNow, if the balloon was large in volume, the change in internal energy ΔE would be positive and the balloon will either burst or move frantically in the refrigerator shelf.\n\nBut, if the balloon was a small one, it'd slowly become limp and deflate as the Q will overpower W (PΔV) in the equation and the internal energy ΔE will decrease and become negative.\n\nPlease note that I have made a lot of assumptions here that perhaps, might not be so easy to apply practically.\n\nThe points you'd have to consider are:-\n\n1-How tightly the balloon is wound so that air doesn't leak out of it.\n2- Whether or not there's a leakage in the balloon.\n3- The temperature of the balloon.\n4- The material of the balloon.\n\nStrictly speaking, the first law isn't even applicable in this case because the condition in the law is that the system (balloon) shouldn't be allowed to move. Its kinetic and potential energies should both be zero which clearly is not the case with our experiment.\n\nI hope this answer helps!\n\nGood luck!
570434
when all parameters r given use ideal gas formula i.e;pv=nrt when temp is kept constant use boyles law ie pv=constant or while solving problems use the form p1.v1=p2.v2 and when pressure is kept constant use charles law and when volume is kept const use pressure temp law morover when u hav rms velocity(=c) given use kinetic gas equation ie pv=mn(c.c)/3 [m=mass of gas molecule,n=total no of gas molecule]
362437
law of conservation of energy\nany gas that is compressed will 'lose' its heat energy to the ambient temperature in the process (ie a refrigerator). in the specific example the gas (1 cm3 at 70F )is at room temperature until its expanded to lets say 1000 cm3, now 70F is spread out 1000x the area,(as is the gas in the can) the more the 70F is expanded the 'colder' it gets (much like the universe or a car tire, but its all relative)
470974
The can and the air inside it expand. The can doesnt expand much, but the air does and a bunch of it flows out. When the can is placed in the cold water the can contracts really quickly and because there is less air pushing back on the walls it crunches up. There is nothing really big happening chemically.
856026
Use the gas law\nPV=nRT
765323
I am not sure what you mean by "General Gas Laws."  The only law dealt with in physics is the Ideal Gas Law which says that pressure and volume are related to amount of gas, temperature and some constant.\n\nPV = nRT
53064
Assuming your gas is an ideal gas, you can use the ideal gas law.\n\nPV=nRT where \nP=Pressure in atm (1 atm= 760 mm Hg so 730 mm Hg= 0.96atm\nV=volume in L\nn= moles\nT=Temperature in K (C=K+273 so 30 C = 303 K)\nR= the universal gas constant= 0.08206 L-atm/mol-K\n\nSolve the ideal gas equation for n=PV/RT\n\nn=[(.96atm)(10L)]/[(.08206 L-atm/mol-K)(303K)]\n\nn=0.386 moles
827958
The first part is correct.  If the numbers in the second part are correct, it is true only for an ideal gas.  Other materials have various densities in their liquid and solid states.
464223
'c' is the right answer.\nuse I law of thermo...\nU=Q-W\nU=Q-Q/4\nU=3Q/4\nBut U=n*C*T where n is moles,C is molar heat cap.for const volume.\nLet us assume that the heat absorbed during the process as\nQ=n* A*T\nwhere A can be defined as molar heat capacity for the process.\nso , n*C*T=3*n*A*T/4\nA=4C/3\nFor a diatomic gas C=5R/2(C=no.of degrees of freedom*R/2)\nA=4*5R/3*2\nA=10R/3