helpppp I just saw my crush when I was walking out of a supermarket and like I was looking a mess so now I'm just wondering how
1 answer:
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A) initial volume
We can calculate the initial volume of the gas by using the ideal gas law:
where
is the initial pressure of the gas
is the initial volume of the gas
is the number of moles
is the gas constant
is the initial temperature of the gas
By re-arranging this equation, we can find:
2) Now the gas cools down to a temperature of
while the pressure is kept constant:
, so we can use again the ideal gas law to find the new volume of the gas
3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:
by using the data we found at point 1) and 2), we find
where the negative sign means the work is done by the surrounding on the gas.
We can calculate the initial volume of the gas by using the ideal gas law:
where
By re-arranging this equation, we can find
2) Now the gas cools down to a temperature of
while the pressure is kept constant:
3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:
by using the data we found at point 1) and 2), we find
where the negative sign means the work is done by the surrounding on the gas.
Refer to the diagram shown below.
i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω
Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5 (10
Also,
R₂*i = 9.5 (2)
Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A
From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)
Answer: 0.08 Ω
i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω
Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5 (10
Also,
R₂*i = 9.5 (2)
Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A
From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)
Answer: 0.08 Ω