Answer:
a) 0 J
b) W = nRTln(Vf/Vi)
c) ΔQ = nRTln(Vf/Vi)
d) ΔQ = W
Explanation:
a) To find the change in the internal energy you use the 1st law of thermodynamics:
Q: heat transfer
W: work done by the gas
The gas is compressed isothermally, then, there is no change in the internal energy and you have
ΔU = 0 J
b) The work is done by the gas, not over the gas.
The work is given by the following formula:
n: moles
R: ideal gas constant
T: constant temperature
Vf: final volume
Vi: initial volume
Vf < Vi, then W < 0 and the work is done on the gas
c) The gas has been compressed. Thus, its temperature increases and heat has been transferred to the gas.
The amount of heat is equal to the work done W
d)
Answer:
No
Explanation:
The reason why no current is produced are basically that, the wavelengths of light in the Balmer transition are reflected, not absorbed in solar panels, hence no current is produced.
The Balmer series consists of lines in the visible spectrum. It corresponds to emission of a photon of light when electrons descend from higher energy levels to the n=2 level in the hydrogen spectrum. The various wavelengths in the Balmer series can be separated by a prism since they are all in the visible region of the electromagnetic spectrum.
In solar panels, light corresponding to the wavelengths in the Balmer series is merely reflected by the panel and not absorbed. Since light is not absorbed, no current can be produced when the panel is irradiated with light corresponding to the wavelengths in the Balmer series.
In mathematics, a percentage is a number or ratio expressed as a fraction of 100.
Answer:
C
Explanation:
a parallel circuit because there is more than one path
Answer:
Uncorrected values for
For circuit P
R = 2.4 ohm
For circuit Q
R = 2.4 ohm
Corrected values
for circuit P
R = 12 OHM
For circuit Q
R = 2.3 ohm
Explanation:
Given data:
Ammeter resistance 0.10 ohms
Resister resistance 3.0 ohms
Voltmeter read 6 volts
ammeter reads 2.5 amp
UNCORRECTED VALUES FOR
1) circuit P
we know that IR =V
2) circuit Q
R = 2.4 ohm as no potential drop across ammeter
CORRECTED VALUES FOR
1) circuit p
IR = V
R= 12 ohm
2) circuit Q
R = 2.3 ohm
