(Question is aimed at Div. C. Sorry Div. B!)
TheChiScientist wrote:Straight outta AP Chem.
When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 °C to 27.5 °C. Calculate the enthalpy change for the reaction in kJ per mol of HCl, assuming that the calorimeter loses only a negligible quantity of heat. The total volume of the solution is 100 mL, its density is 1.0 g/mL, and its specific heat is 4.18 J/g-K.
 \cdot 1.0\ \frac{\textrm{g}}{\textrm{mL}} \cdot 100\ \textrm{mL} \cdot \frac{1\ \textrm{kJ}}{1000\ \textrm{J}}=2.717\ \textrm{kJ})
(Since we're considering a range, actually converting the temperatures to Kelvin would yield the same answer.)
Technically, the final answer should have one significant figure, but...
Nice. AP Chem HW done!UTF-8 U+6211 U+662F wrote:TheChiScientist wrote:Straight outta AP Chem.
When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 °C to 27.5 °C. Calculate the enthalpy change for the reaction in kJ per mol of HCl, assuming that the calorimeter loses only a negligible quantity of heat. The total volume of the solution is 100 mL, its density is 1.0 g/mL, and its specific heat is 4.18 J/g-K.Side note: I wish the built-in
(Since we're considering a range, actually converting the temperatures to Kelvin would yield the same answer.)
Technically, the final answer should have one significant figure, but...renderer included the SI package...
(Jk this was due a week ago. 
)5 times, 0, free expansion is irreversible and the entropy change is greater than zero, while reversible expansion has no entropy change.UTF-8 U+6211 U+662F wrote:A gas is allowed to freely expand to a box which is five times the volume the previous container was. What is the resulting entropy change? A gas expands reversibly until its volume is five times as much. What is the resulting entropy change? Explain the difference between free expansion and reversible expansion.
MattChina wrote:5 times, 0, free expansion is irreversible and the entropy change is greater than zero, while reversible expansion has no entropy change.UTF-8 U+6211 U+662F wrote:A gas is allowed to freely expand to a box which is five times the volume the previous container was. What is the resulting entropy change? A gas expands reversibly until its volume is five times as much. What is the resulting entropy change? Explain the difference between free expansion and reversible expansion.
1) Increases by [math]Nk_B\ln 5[/math] or equivalently [math]nR\ln 5[/math], where N is the number of particles, k is the Boltzmann constant, n is the number of moles, and R is the ideal gas constant 2) Yep! Just note that only the net entropy change (of both the gas and the surroundings) is 0. The entropy of the gas itself can increase or decrease. 3) Yep!
Clausius's formulation was "it is impossible for a self-acting, unaided by any external agency, to trader heat from a body at lower temperature to a body, at a high temperature.UTF-8 U+6211 U+662F wrote:What were Clausius's and Kelvin's formulations of the Second Law of Thermodynamics?
Yep, your turn.Crimesolver wrote:Clausius's formulation was "it is impossible for a self-acting, unaided by any external agency, to trader heat from a body at lower temperature to a body, at a high temperature.UTF-8 U+6211 U+662F wrote:What were Clausius's and Kelvin's formulations of the Second Law of Thermodynamics?
Kelvin's formulation was "it is impossible to obtain a continuous supply of energy by cooling a body below the coldest of it's surroundings."
