Chapter 6: Problem 10
In the equation \(w=-P \Delta V,\) why is there a negative sign?
Chapter 6: Problem 10
In the equation \(w=-P \Delta V,\) why is there a negative sign?
All the tools & learning materials you need for study success - in one app.
Get started for freeWhich has the greater kinetic energy, an object with a mass of 2.0 \(\mathrm{kg}\) and a velocity of 1.0 \(\mathrm{m} / \mathrm{s}\) or an object with a mass of 1.0 \(\mathrm{kg}\) and a velocity of 2.0 \(\mathrm{m} / \mathrm{s}\) ?
A piston performs work of \(210 . \mathrm{L} \cdot\) atm on the surroundings, while the cylinder in which it is placed expands from \(10 . \mathrm{L}\) to 25 \(\mathrm{L}\) . At the same time, 45 \(\mathrm{J}\) of heat is transferred from the surroundings to the system. Against what pressure was the piston working?
Quinone is an important type of molecule that is involved in photosynthesis. The transport of electrons mediated by quinone in certain enzymes allows plants to take water, carbon dioxide, and the energy of sunlight to create glucose. A 0.1964 -g sample of quinone \(\left(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}\right)\) is burned in a bomb calorimeter with a heat capacity of 1.56 \(\mathrm{kJ} / \mathrm{C}\) . The temperature of the calorimeter increases by \(3.2^{\circ} \mathrm{C}\) . Calculate the energy of combustion of quinone per gram and per mole.
Calculate the internal energy change for each of the following. a. One hundred \((100 .)\) joules of work is required to compress a gas. At the same time, the gas releases 23 \(\mathrm{J}\) of heat. b. A piston is compressed from a volume of 8.30 \(\mathrm{L}\) to 2.80 \(\mathrm{L}\) against a constant pressure of 1.90 \(\mathrm{atm}\) . In the process, there is a heat gain by the system of 350. J. c. A piston expands against 1.00 atm of pressure from 11.2 \(\mathrm{L}\) to 29.1 \(\mathrm{L}\) . In the process, 1037 \(\mathrm{J}\) of heat is absorbed.
The preparation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) is an endothermic reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)(\text { unbalanced }) $$ The enthalpy change of reaction for the balanced equation (with lowest whole- number coefficients) is \(\Delta H=67.7 \mathrm{kJ}\) . If \(2.50 \times 10^{2} \mathrm{mL} \mathrm{N}_{2}(g)\) at \(100 .^{\circ} \mathrm{C}\) and 3.50 atm and \(4.50 \times\) \(10^{2} \mathrm{mL} \mathrm{O}_{2}(g)\) at \(100 .^{\circ} \mathrm{C}\) and 3.50 atm are mixed, what amount of heat is necessary to synthesize the maximum yield of \(\mathrm{NO}_{2}(g) ?\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.