resistors

1. 45 A of current passes through a cross section of wire. How many charges pass through in 3 seconds? A. 9.375 x 10^19 B. 3.59 x 10^16 C. 2.16 x 10^17 D. 8.4 x 10^20 2. A 30 ohm resistor dissipates 500W of power. What current is running through the resistor? A. 15000 A B. 16.6 A C. 4.06 A D. 0.052 A 3. How is current affected if the number of charge carriers decreases? A. The current increases. B. The current decreases. C. The current initially decreases and then is gradually restored. D. The current is not affected. 4. The current in an electron beam in a cathode-ra... click for more

Macroscopic Description of an Ideal Gas

A rigid tank contains 0.40 mol of oxygen (02). Determine the mass (in kg) of oxygen that must be withdrawn from the tank to lower the pressure of the gas from 40 atm to 25 atm. Assume that the volume of the tank and the temperature of the oxygen are constant during this operation. This is how I solved it. I just want to confirm that my answer is correct. Pi/ni = Pf/nf 40/0.40mol = 25/n n = 0.25 mol 0.25 mol O2 * (32g O2)/1 mol O2 * (1*10-3 kg)/1 g O2 = 0.008 kg O2 Is this correct? Thanks

Molar heat capacity at constant volume and pressure

When 117 J of energy is supplied as heat to 2.00 moles of an ideal gas at constant pressure, the temperature rises by 2.00 K. calculate the molar heat capacity at constant pressure and the molar heat capacity at constant volume for the gas. is the gas monatomic or diatomic?

Adiabatic condition and equation of state of an ideal gas

A sample of 8.02 x 10^-1 moles of nitrogen gas (gamma=1.40) occupies a volume of 2.00 x 10^-2 m^3 at a pressure of 1.00 x 10^5 Pa and temperature 300K. the sample is adiabatically compressed to half its original volume, nitrogen behaves as an ideal gas under these conditions. Show from the adiabatic condition and the equation of state that TVgamma-1 remains constant, and hence determine the final temperature of the gas? calculate the change in internal energy of the gas and the work done in compressing the gas?

Laws of Thermodynamics

Question: When a gas follows path 123 on the PV diagram in the attached file, 418 J of energy flows into the system by heat, and -167 J of work is done on the gas. a) What is the change in the internal energy of the system? I believe that I have this part correct. Change in U = Q + W = 418 J + -167J = 251 J b) How much energy Q flows into the system if the gas follows path 143? See attached file for diagram. The work done on the gas along this path is -63.0 J. Based on this information, I'm not sure how to solve for Q. The only relationship I know of is: change in U = Q + W. I can't use the U I found in part A right? A... click for more