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Q=0 in an adiobatic system because it's an insulated system (let's imagine an ideal system) where there's no temperature change We need to consider case under ideal gas equation #pv=nrt# where #p#, is the pressure, #v# is the volume, #n# is the number of moles, #t# is the temperature of the gas in kelvin. If change in internal energy is always equal to #n*cv*dt# and there's no temperature change, then there shouldn't be a change in internal energy
But this wouldn't make sense because the system would overall just be undefined or something. We see that from state 1 to 2 both pressure and volume are changing By the same logic, it will eventually become spontaneous as temperature increases
To test that, take the limit value #deltag^@ = 0# and find the minimum temperature that will allow for that to happen
The heat is proportional to the motion of the molecules of an object The heat of an object is the total kinetic energy of the molecules (or atoms) of that object Therefore the more heat an object has the more it's molecules are moving The difference between heat and temperature is that temperature is the average amount of kinetic energy the particles of an object has, whereas the heat is.
You can't find the energy with just the information you have given More information is required but here are a couple of possibilities 1) if you are submerging an object in a constant temperature water bath for two hours, the object will presumably reach a temperature that is the same as the water bath (thermal equilibrium) If you know this value (say 60^oc), the starting temperature (say.
Neither of the previous answers have defined a saturated solution, and would properly be rejected by an examiner
Saturation defines an equilibrium condition A saturated solution is a solution in which the concentration of the solute is equal to that concentration that would be in equilibrium with undissolved solute As for any equilibrium we would normally specify a temperature (because at. The gas molecules have some amount of kinetic energy by virtue of which they can move randomly.but when the temperature of the gas is decreased or pressure on the gas is increased, the gas becomes liquid at a certain temperature or pressure.here when we decrease temperature or increase pressure the randomness of the gas molecules are decreased & consequently the kinetic energy of.
The first thing to do here is determine how much energy in the form of heat is required to increase the temperature of that sample of coffee from #25.0^@c# to #62.0^@c#. 154500cal we know that heat gained/lost is given by deltaq=mst, or deltaq=ml where m,s and t are the mass, specific heat and rise or gain in temperature of the object L is the latent heat for the change of state In the given problem heat is given to water to increase its temperature from 22^@c to 100^@c and thereafter boil the water to make it steam at 100^@c
The final temperature is t_2 =806.9 k using the ideal gas law, pv=nrt, and assuming the volume is being held constant, you will have p_1/ (n_1*t_1) = p_2/ (n_2*t_2) however, the number of moles is constant as well, given the fact that the gas is simply heated, so the ecuation becomes p_1/t_1 = p_2/t_2 this will result in t_2=p_2/p_1 * t_1 = 0.880/0.325 * 298=806.9 k keep in mind the.
From state 1 to state 2 it is given that the #pv# graph has been obtained while given mass of gas was subjected to temperature changes
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