Workdone by compressing an ideal gas3/19/2023 100 liters of an ideal gas at 25.0 C is placed in a closed, adiabatic container that is kept at a constant pressure of 1.00 atm.A very small change from the reaction enthalpy at 25 C. So the reaction enthalpy at 50 C on a molar basis is -242.0 kJ/mole. The last term is heating the H 2O from 25 to 50 C (which is positive because this requires heat from the environment). The next term is the reaction enthalpy at 25 C given in the problem for one mole of water formed. The first two terms are for cooling one mole of H 2 and half a mole of O 2 from 50 to 25 C (they are negative because this is cooling and heat is leaving the system). So we start with 50 C hydrogen and oxygen, cool them to 25 C, undergo the reaction at that temperature, then heat the H2O gas back up to 50 C.ĭH R,50 = -C P,H2 D T - (1/2)C P,O2 D T + DH R,25 + C P,H2O D T Here the strategy is to realize that enthalpy is a state function. State your answer for this problem to 4 significant digits. If the molar heat capacities at constant pressure of H 2, O 2 and H 2O(gas) are 28.82, 29.36 and 33.58 J/(mole K), respectively, what would the enthalpy of reaction be at 50.0 C? Assume that the gases all behave ideally before and after the reaction occurs. H 2O, (all in the gas phase) the standard enthalpy of reaction at 25.0 C is -241.8 kJ/mole (-241.8 kJ per mole of water formed). Since the system is isothermal and an ideal gas, we immediately can write down thatĭ H = D U = 0. Determine q, w, DU and DH for the process. 0.500 moles of an ideal gas is expanded isothermally and reversibly from 10.0 liters to 20.0 liters at a temperature of 30.0 C.Numerical Problems (each problem is worth 19 points). Therefore, the change in enthalpy will be greater since for an ideal gas C P = C V + nR. The change in enthalpy will be given by C P times the change in temperature. The change in internal energy will just be given by C V times the change in temperature. Which changes more, the internal energy or the enthalpy? Explain. One mole of an ideal gas is taken from a state in which P = 1atm, T= 25 C, and V = 1 liter to a state in which P = 1.2 atm, T = 192 C and V= 1.3 liters.The change in internal energy of any closed adiabatic system at constant volume is zero (assuming that only PV work is possible). Burning gasoline inside a closed, adiabatic system at constant volume will cause the internal energy to increase, decrease or stay the same? Explain.This means q = -w and for a compression, w is positive. Since the expansion is isothermal and of an ideal gas, the change in internal energy is zero. For the isothermal, reversible compression of an ideal gas, will the heat (q) be positive, negative or zero? Explain.The internal energy will increase more if the system is at constant volume because at constant pressure, some of the energy input as heat goes into work done on the surroundings instead of all going into the kinetic energy of the gas molecules. Assume that the constant pressure system is surrounded by an environment at 1 atm. Will the internal energy increase more if the system is at constant volume or at constant pressure? Explain. You must give a brief explanation of your answer for full credit (a sentence or an appropriate mathematical expression). You should only have your exam, writing implements and a calculator on your desk.Ĭonceptual Problems (each problem is worth 6 points). The last page of the exam is a list of constants and equations. Count them to insure that they are all there. The exam consists of 6 pages, including this one. ID#_ĭo not open this exam until told to do so.
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