Explain the effect of heating a fixed mass of gas at constant volume.?
960494
Using the Ideal Gas Law, PV=nRT . . .\n\nR is the Gas Constant, and doesn't change.  If you don't change the number of moles of gas (n), and volume is constant (V1 = V2), then you simplify the equation to:\n\nP/T = nR/V = constant\n\ntherefore, any increase in temperature, at constant volume and number of moles, must result in a proportional increase in pressure, since the ratio of pressuredivided by temperature is a constant.
751958
heat change\ncolor change\nemmission of gas
902070
In weight, it's the same. No mass has been added or taken away by heating it or cooling it down, weight doesn't care about tempurature except possibly at phase changes.\n\nIt does however become more dense until about 4 degrees centigrade, so water of a certain volume at 4 degrees centigrade ways more then water of the same volume at 10 degrees centrigrade. This is because the 4 degree centigrade water is more dense so to get the same volume you need to add more mass.
606777
the actual equation is:  (p + b·n^2/V^2)(V - a·n) = nRT\n\na:- every mol of particles of the gas has a tiny finite volume of a\n\nb:-modification is due to the particles sticking to each other, a phenomenon called cohesion\n\nR- rydberg's constant
378824
Temperature affects state change. If something, for instance water, is heated, it becomes a gas. Gases take up more volume than liquids. So the hotter a gas is, the larger it will expand, and the same is true, to a lesser degree, for solids and liquids. Hope I helped,\nZ
358671
Gay-Lussac's Law states that for a fixed amount of gas (the number of moles stays constant throughout the problem) and a fixed (constant) volume, pressure and temperature are proportional, the formula is:\n\nP1/T1=P2/T2  \n\nThis means as temperature increases in a fixed volume, the gas exerts more pressure on the container, they expand and hit the container walls and push on them.  If the gas is cooled and kept in the same container, the particles exert less pressure on the container, they're more condensed.\nTemperature must be in Kelvin (K) not Celcius to use the formula.
542119
As described by the ideal gass laws, P1 V1 = P2 V2. This means that, as the pressure increases, the voulume of the contained vapor decreases and condences.
354017
its a high pressure heat with constant movement that create volume, density conductors
1128521
gas cools down if it expands suddenly.\n\nsimilar principle is used in AC and refrigerators.
501359
yeaaaaah\n\nit's "expand and rise"\nthe energy (heat) supplied to it causes the individual molecules to have more energy and because of it, they expand further, and rise since density is lowered and volume is increased
1080952
mass, volume and temperature.
755161
The heat energy Q  that must be absorbed by a mass m with specific heat c raises the temperature of the mass from temperature T1 to temperature T2: \n\nQ = mc(T2-T1)\n\nThis assumes there are no other energy losses or gains to the system.
1118780
Your question is very open.\n\nheat equations very general can be:\napplied heat = specific heat x mass x increase of temperature\napplied heat = heat capacity x increase of temperature\napplied heat = mass x specific melt heat\napplied heat = mass x specific evaporation heat (at boiling point)\nHeat from a supply = power x time.\nCombinations and others are possible.\n\nwave equation can be x = A sin( 2pi t/T)\nThere are more possibilities.
675016
This is a calorimetry problem! There is an equation for it:       \nq = mc(change in temperature)\n\nq = the heat required \nm= mass\nc= specific heat \n\nAlso, write out the reaction and products.
131636
In pretty much all substances, when you heat something (assuming no phase change), the material becomes less dense due to the fact that its volume expands and mass stays constant.\n\nSo if you were to heat a pot of water for example from the bottom on a convection stove, the hot water at the bottom would become less dense than the cooler water on top.  The hot water would rise and it is rises, the cooler water sinks down to take its place.
569322
Use the ideal gas law \n\nPV/T = constant for the final and initial states.
801336
If you neglect the loss in mass due to the energy given off in the process (a very, very, tiny amount of mass at that), then the total mass should stay constant before and after the reaction.\n\nIf you have the reaction,\nA --> B\nThe mass of A equals the mass of B\n\nHowever, that is not to say that what is leftover after burning has the same mass as the original substance burned.\nIn a standard combustion reaction, two products are produced, Carbon Dioxide and water.  Both of these products are in the gas phase and "float" away, taking with them part of the mass of the original substance.  If one could trap and collect all the particles of a reaction in an enclosed space, then the mass will be constant, but in an open system where things can escape, you will generally loose part of the mass in this manner.