A tank containing 200 L of hydrogen gas at 0.0 Celsius is kept at 10 kPa. The pressure is raised to 95C, and the volume is decre
1 answer:
Answer:
The new pressure of the gas is 15.40 kPa.
Explanation:
Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. Mathematically this law indicates that the quotient between pressure and temperature is constant:
On the other hand, Boyle's law says that the volume occupied by a certain gaseous mass at constant temperature is inversely proportional to the pressure. This law is expressed mathematically as:
P*V=k
Finally, Charles's law indicates that as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. Mathematically, this law says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:
Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:
Studying an initial state 1 and a final state 2, it is fulfilled:
In this case:
- P1= 10 kPa
- V1= 200 L
- T1= 0 C= 273 K
- P2=?
- V2= 175 L
- T2= 95 C= 368 K
Replacing:
Solving:
P2= 15.40 kPa
<u><em>The new pressure of the gas is 15.40 kPa.</em></u>
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Answer:
Q = +1.4 pC
Explanation:
- The capacitance of any capacitor, by definition, is as follows:
-
- Applying Gauss'Law and the definition of electric potential, it can be showed that the capacitance of a parallel-plate capacitor can be expressed as follows:
-
- Where εr is the dielectric constant of the material that fills the space between the plates, A is the area of one of the plates, and d, is the separation between them.
- Replacing by the givens in (2) we can find the value of the capacitance C, as follows:
- Replacing the values of C and V in (1), we can solve for Q, as follows:
- As the outer surface is at a higher potential that the inside surface, the charge on it must be positive, and is equal to +1.4 pC.
Answer:
W = 2.3 10²
Explanation:
The force of the weight is
W = m g
let's use the concept of density
ρ= m / v
the volume of a sphere is
V = π r³
V = π (1.0 10⁻³)³
V = 4.1887 10⁻⁹ m³
the density of water ρ = 1000 kg / m³
m = ρ V
m = 1000 4.1887 10⁻⁹
m = 4.1887 10⁻⁶ kg
therefore the out of gravity is
W = 4.1887 10⁻⁶ 9.8
W = 41.05 10⁻⁶ N
now let's look for the electric force
F_e = q E
F_e = 12 10⁻¹² 15000
F_e = 1.8 10⁻⁷ N
the relationship between these two quantities is
= 41.05 10⁻⁶ / 1.8 10⁻⁷
\frac{W}{F_e} = 2,281 10²
W = 2.3 10²
therefore the weight of the drop is much greater than the electric force