Describe what the ideal gas law means (no equations)
The ideal gas law describes the relationship between the "state variables" temperature, pressure, and volume. Pressure and volume are directly related to temperature and number of moles (molecules)while pressure is inversely related to volume. They are related by a constant of proportionality (either R or k).
Calculate the mass of a mole of dry air, which is a mixture of N2 (78% by volume), O2 (21%), and argon (1%). (Hint: use periodic table)
29 g
Compare and contrast heat and work
The both are forms of energy transfer. Heat transfer occurs due to a temperature difference. Work is any other type of energy transfer in a system.
Problem 1.45 parts (a)+(b)
(a) w=(y^2)(z) and w=(x2)/(z)
(b) (dw/dx)y = y
(dw/dx)z = 2x/z
Describe what the first law of thermodynamics means (no equations)
By applying a pressure of 200 atm, you can compress water to 99% of its usual volume. Sketch this process (not necessarily to scale) on a PV diagram, and estimate the work required to compress a liter of water by this amount. (hint: a joule is = (atm)(L))
100 J
A battery is connected in series to a resistor, which is immersed in water (to prepare a nice hot cup of tea). Would you classify the flow of energy from the battery to the resistor as heat or work? What about the flow of energy from the resistor to the water?
Energy from the battery to the resistor is work
Energy from the resistor to the water is heat
In a Diesel engine, atmospheric air is quickly compressed to about 1/20 of its original volume. Estimate the temperature of the air after compression, and explain why a Diesel engine does not require spark plugs. (Hint: this is an adiabatic process)
~1000 K
Describe what the enthalpy of an material means (no equations)
Enthalpy is the total energy of a system plus the work necessary to make room for that system
Problem 1.51: Use the data at the back of this book to determine the change in enthalpy for the combustion of a mole of glucose,
C6H12O6 + 6O2 ----> 6CO2 + 6H2O
dHglucose = +1273 kJ (negative enthalpy of formation of glucose)
dHco2+H20 = 6(-393.5 kJ) + 6(-285.8 kJ) = -4076 kJ
dHtotal = -2803 kJ