Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid-vapor mixture at a pressure of 200 kPa. The refrigerant absorbs 120 kJ of heat from the cooled space, which is maintained at -5°C, and leaves as saturated vapor at the same pressure. Determine (a) the entropy change of the refrigerant, (b) the entropy change of the cooled space, and (c) the total entropy change for this process.  Please help explain to me how to do this!!! I don't know where to start... any help would be great!
133705
Take table for R134a refrigerant:\n\na) delta s = entropy change\n--inlet for evaporator assuming 50%moisture= interpolate to get S \n@200kpa for R134a\n0       Sg   \n0.5     S\n1       Sf\nget Sg and Sf from table at 200kpa for R134a. \n \n--out for evaporator = Sg at 200Kpa (get from table) since it is sat vap.\n\nthe difference of both Sfg and Sg is the answer.\n\nb)  delta S = Q / T  where Q = 120 KJ and T = -5+273 = 268K\n\nc) add both a) and b) is the total entropy
74026
The guy who answered before was right but i wanna explain it. An exothermice reaction is one in which heat energy is lost to the surroundings, thus at the end of the reaction, some heat would have been lost. For heat to have been lost, the product has to have lesser heat than the reactant.Take not that heat here is the same as Enthalphy.\nSince the Enthalphy heat value = heat of product-heat of reation = a negative number because heat of product is lesser.\nthus the answer is i nd ii\n\n2. To calculate the heat absorbed or evolved in a system, the formulais Q(quantity of heat) = m*c*T, where m is the mass, c is the specific heat capacity, and T is the change in temperature, thuis this three are neede. Density is not important in this calculation\nso the answwers are i and iii
471505
Heat released = 480*2.34 J\n\nSo 0.02 mol releases this much \n\n1 mol will release 480*2.34/0.02 = 56.16 kJ
665998
While I was reading the question, my wife asked me to buy 5 liters of kerosene urgently.\n\nI went to the shop; the vendor had kerosene in a barrel; I asked him to empty 5 liter of kerosene from his barrel.  He poured 5 liter of kerosene in my 10 liter can. \n\nHe told that he has emptied 5 liter of air from the can.  I paid him for having emptied 5 liter of air from the can. I also checked whether 5 liter of air is filled up in the vendor’s barrel.\n\nComing home I saw the answer page ¾ empty. To my part I thought to fill the page with some answer. This is the result.
419734
Your experiment was a demonstration of the ideal gas law, which establishes a specific relationship between the temperature, pressure, and volume of a gas.  Let me explain:\n\n1) When you boiled the flask of water and then plugged the top, the temperature of the gas trapped in the top of the flask would have been approximately the normal boiling point of water, 100 degrees C.\n\n2) By plugging the top of the flask, you are, in effect, creating a small system inside the flask where the volume of the gas is held constant -- no gas goes in or out.  In this situation, according to the ideal gas law, the pressure and the temperature of the gas will increase or decrease together.  So, when you flip the flask over and place the bag of ice on its surface, the ice lowers the temperature of the gas inside, which also lowers the pressure of the gas, according to the law.\n\n3) As the pressure of the gas decreases, this in turn decreases the boiling point of the water inside.  Why?  Because a liquid boils when its 'vaporization pressure' equals the pressure of the atmosphere around it.  In other words, as a liquid gets hotter, the individual atoms or molecules move around with increasing force, and at some point they begin to break free of the liquid -- this tendency for particles to break free is the vaporization pressure, and obviously it increases as the temperature of the liquid goes up.  At the boiling point, the vaporization pressure matches the pressure that the surrounding atmosphere is exerting on the liquid, and at this point particles begin to break free and vaporize all throughout the liquid. \n\nIn your experiment, the water in the flask had cooled a few degrees below its boiling point under normal atmospheric pressure (100 C).  However, when the ice lowers both the temperature and the pressure of the gas (the atmosphere) surrounding the water, all of a sudden the atmospheric pressure drops below the vaporization pressure of the water (which is still very high, since water cools slowly), and so the water boils again, at a temperature less than 100 C.  Interesting stuff.\n\nHope this answer isn't too long.  I wasn't sure how much you'd like explained.
184833
R is the gas constant (= 8.31 J/mol*K).  The value should be with the list of constants in your book.\n\nOk Nic, you seem to be a bit confused on this issue so lets go back and start from scratch.\n\nFirst you obtained you values for T and K (experimentally or otherwise).  MAKE SURE YOU CONVERT YOUR TEMP VALUES TO KELVIN.  It's important for the units of the constant (R) are in Kelvin.  So then you calculate 1/T and lnK and plot the values for each set to generate your graph.\n\nYou calculate slope by taking two of your points (x1,y1) and (x2,y2).  From the data you added, let's say we'll use your first (0.002976, -3.079) and last points (0.003165, -4.225).\n\nslope= [-4.225- (-3.079)]/[0.003165-0.002976]= -6063.49 kelvin\n\nSince slope = -delta H/R\nThen -delta H = slope * R\nSo -delta H = -6063.49kelvin * 8.31 J/molkelvin\nAnd -delta H= -50387.62 J/mol\nRemoving the negatives from both sides\nDelta H= 50387.62 J/mol\n\nFor the second part of the question, use one (x,y) experimental point to find the y-intercept.  They don't want you to determine it from your plot, they want you to calculate the exact value.\n\nThe equation of a line is y=mx+b where "m" is the slope and "b" is the intercept.  So take another experimental point from readings (ie 0.003012, -3.751) and calculate the intercept by using the slope value calculated above.\n\nIf y=mx+b, then b= y-mx\nPlugging in the values\nb= -3.751 - (-6063.49kelvin * 0.003012/kelvin)\nb= -3.751 - (-18.26)\nb= 14.51\n\nTo calculate delta S for that same point (for that's what is requested), first you have to calculate delta G for that point.  Use the  following equation;\n\ndelta G = -RTlnK\n\nYou know the value for R (8.31 J/molkelvin), and you have the temp value and lnK value for that point for it was expermientally determined\n\n1/T= 0.003012/kelvin so T=332.01 kelvin\nlnK= -3.751\n\nSo delta G= -8.31 J/molkelvin * 332.01 kelvin * -3.751\nand delta G = 10349.02 J/mol\n\nYou can then calculate delta S for\nintercept = delta S/ delta G (given in your original data)\n\nTherefore delta S = intercept * delta G\nSo delta S = 14.51 * 10349.02 J/mol = 150164.28 J/mol\n\nHope this clarifies things.
416534
The question is if you start with gas at 20.0 atm and 20.3 C and heat it to 125 C what will be the pressure?  It is a changing condition problem making use of Charles' Law.  The answer is 27.1 atm.
658154
I would visit the site cited below.  The discussion includes:\n\n"Irving Langmuir developed a way to measure vapor pressure \nby measuring the evaporation rate, which we will now flip around backwards. His reasoning is that the rate at which molecues are lost due to evaporation to a gas with no partial pressure of the evaporating substance is the same as the rate at which molecules of the substance would hit the surface if it were in equilibrium with the vapor (because in equilibrium the evaporation rate and re-condensation rate cancel each other out). His expression is: \n\n(mass loss rate)/(unit area) = (vapor pressure - ambient partial pressure)*sqrt( (molecular weight)/(2*pi*R*T) ) "
533155
First, you have to understand what an ideal gas is. Read your book.\n\nSecond, derive your ideal gas equation to get V for Q1, and P for Q2. This is easy.\n\nTo convert Celsius to Kelvin, add 273 to the temperature you have in Celsius. So, if you have 17C, its value in Kelvin is 290K.\n\nFor the gas constant, \nR = 0.08206 L*atm / mol*K\n\nPlug your values to the ideal gas equation. Good luck!
397112
The formula would be 2HCL+ CaCO3 ---> CaCl2 + H2O + CO2. The standard enthalpies of formation are -167.2, -1207, -796, -285.8, and -393.5 respectively. When you use the formula "Sum of products minus sum of reactants, you come up with +66.1 (don't forget to multiply the HCl formation by 2 for 2 moles). I am assuming that the HCl you are talking about is HCl (aq) and not HCl (g). If it is a gas, then the standard enthalpy of formation is -92.31, so you would come up with a different number (-83.68)\n\nIf it is aqueous HCl, you come up with a positive 66.1 kJ/mole, which means that heat was absorbed by the reaction (endothermic). That probably means your cylinder would feel cooler, since heat is being taken in by the reaciton.\n\nIf you mean gaseous HCl, you come up with a negative 83.68 kJ/mol, which means that heat was given off by the reaction (exothermic). That probably means yoru cylinder woudl feel warmer, since heat is being given off by the reaciton and taken in by the cylinder.
502015
all u have to do before attempting algebra is stop panicking.\nBe cool and try to substitute unknown with some variable say x.\nSo the radiator has a capacity of 16 QT.\nAnd you need 40% antifreeze solution.\nSo u need (40*16)/100 quart of pure antifreeze\nwhich is equal to 6.4 pure antifreeze.\nSo suppose u need to drain x quart of antifreeze solution already in the radiator. You replace the drained solution with pure antifreeze solution.\nso form a equation:\n25% of (16-x)+ x= 6.4 \n25*(16-x)/100 + x = 6.4\n(16-x)/4 + x= 6.4\n16-x+4x=4*6.4\n3x = 25.6- 16\nx=9.6/3\nx=3.2 quart\nso u drain 3.2 quart of 25% antifreeze solution originally in the radiator and replace it with equal amount of pure antifreeze.\nHope this helps
453043
2.81 L. convert the temperatures to Kelvins and the 180. torr to atm. then plug it in into the Combined Gas Law.
509764
You must approach this question with an understanding of energy conservation. Change in Energy = mass*heat capacity*change in temperature, where heat capacity = 1.00 for pure water. By evaluating your conditions with the above definition of thermal energy, it will show that you must supply 2 calories of heat to the system.\n\nIf you were to add another substance to the original system. You would once again approach the problem using energy conservation. The energy of a hot liquid (mass1*heat capacity*change in temperature) + the energy of a cold liquid (mass2*heat capacity*change in temperature) = mass(1+2)*heat capacity*change in temperature.\n\n-Christopher
428617
At constant temperature and pressure, a reaction will occur spontaneously if the Gibbs free energy change of the reaction is less than zero.\n\nThe Gibbs free energy change, delta-G, is given by:\n\ndelta-G = delta-H - T*delta-S\n\nwhere T is the absolute temperature, \n\ndelta-H is the enthalpy change for the reaction at the temperature and pressure of interest, and \n\ndelta-S is the entropy change for the reaction at the temperature and pressure of interest.\n\nWe are given the enthalpy and entropy changes for the reaction of interest in this question.  I will assume that the enthalpy change and entropy change for this reaction are independent of temperature. Note also that the values of both of these thermodynamic parameters are given in "per mole" units in this question.  I will assume these values are normalized to one mole of one of the reactants, and not one mole of the product.\n\nPlugging in the enthalpy and entropy values, and setting delta-G to be less than zero:\n\n0 > delta-G = 180500 J/mol - T*24.8 J/(mol*K)\n\nT  > (180500 J/mol)/(24.8 J/(mol*K))\n\nT > 7278 K\n\nThis is an extremely high temperature, and the assumption that we made above that the enthalpy and entropy changes were independent of temperature is probably not very good given the large temperature extrapolation being made here.
55868
The original water temp is 25C, as was given in the problem... I'm assuming you meant to ask what was the initial temp of the water that was added, which isn't too hard. Here goes.\nQ(heat)= m(mass) x c(specific heat) x delta T(change in temp)\nso Q = (150g)(4.something J/gC-look it up)(25C-14C)\nBecause of conservation of energy and assuming ideal insulation, the heat LOST by the 150g of water during the mixing is equal to the heat GAINED by the 200g of water. So, once you have Q, rearrange the equation to look like [delta T] = Q/mc = Q/(200g)(c)\nThat will give you the temperature difference between 14C and the 200g of water's initial temp. Delta T should come out as being positive, since the temp is going up. \nWhen you have more complex mixtures of things with different specific heats and add processes and changes, the math becomes a lot more of a headache but the same basic principles apply just the same.
238543
Information on absorption refrigeration at this site:\nhttp://www.nicorinc.com/en_us/commercial/products_and_services/air_quality/absorption_works.htm
140880
Cooling, Heating, and Dehumidifying\n\n3in1Fig1.jpg (23385 bytes)\n\n    As you turn on the unit,  the refrigerant in the system (in liquid form and with low temperature and low pressure), pressurized by the compressor, turns into gas. The gas then enters into the condenser and perform a process called heat exchange. In this process, refrigerant turns back into liquid by releasing heat energy to the air (Airflow 2) and becomes cooler than before. The liquid refrigerant with cooler temperature enters into the evaporator and perform another heat exchange with the air (Airflow 1). In this process, the heat energy in the air is absorbed by the refrigerant in the evaporator and the resulting cooler air is exhaust through the front vent. Note: Airflow 1 and Airflow 2 are independent of each other.\n\n     \n\n    Because the evaporator is lower in temperature than the air, water vapor will condense if the evaporator's temperature is lower than the ambient dew point. This cycle of condensation will lower the Relative Humidity of the room and thereby achieve the goal of dehumidification.\n\n    3in1Fig3.jpg (16322 bytes)  3in1Fig4.jpg (19482 bytes)\n\n    However, if the unit is used as a dehumidifier, remember to remove the air exhaust duct from the back in order to effectively dehumidify the room. If not, constant air flow from the outside (in order to compensate the air flow to the outside through the air exhaust duct) will make the dehumidifying process ineffective.\n\n     \n\n    Heating function (optional in most models) is included with the intention to make this unit an all-season air-conditioning unit. Operationally, this function is just the same as other features. You can operate the fan, louver, and timer, just as you can for cooling.\n\n________________________________________________________________