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Anonymous users2024-02-05The law states that the pressure [1] of a gas of a certain mass, when its volume is constant, is proportional to its thermodynamic temperature. Namely.
P1 P2 T1 T2 or Pt P 0 (1 t.)
where p 0 is the pressure of the gas when 0 is 0 and t is the temperature in Celsius.
The pressure p and temperature of the gas can be easily determined with a piezometer and a thermometer. The law is verified if the P t plot line is a straight line through the coordinate origin, or if the p t plot line is a straight line and intersects the t axis at minus 273 degrees Celsius.
Note: It is absolute zero, which can only be approximated infinitely, and cannot really be reached).
The law states that the pressure [1] of a gas of a certain mass, when its volume is constant, is proportional to its thermodynamic temperature. Namely.
P1 P2 T1 T2 or Pt P 0 (1 t.)
where p 0 is the pressure of the gas when 0 is 0 and t is the temperature in Celsius.
The pressure p and temperature of the gas can be easily determined with a piezometer and a thermometer. The law is verified if the P t plot line is a straight line through the coordinate origin, or if the p t plot line is a straight line and intersects the t axis at minus 273 degrees Celsius.
Note: It is absolute zero, which can only be approximated infinitely, and cannot really be reached).
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Anonymous users2024-02-04Isn't it interesting to copy around?
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Anonymous users2024-02-03t1300 k…①
When the piston has no pressure on the ground to the piston: p1
s+mg=p0s, i.e., p1
p0mgspa…When the temperature rises to t2
When the cylinder has no pressure on the ground, for the cylinder: P2
s=p0s+mg
Immediate excitation: P2P0mg
SPA to the gas in the cylinder: isochoric change, obtained by Charlie's law: pt
pt simultaneous solution yields: t2
400k i.e.: t2
Answer: When the gas temperature in the sock cylinder rises to 127 degrees Celsius, the cylinder has no pressure on the ground shielding surface
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Anonymous users2024-02-02（1）t1
300 k…①
When the piston has no pressure on the ground to the piston: p1
s+mg=p0
s stands for: pp?mgs
Pa(2) when the temperature rises to T2
When the cylinder has no pressure on the ground, for the cylinder: P2
s=p0s+mg
i.e.: P2P+MGS
Pa vs. gas in the cylinder: isovolume change, obtained by Charlie's law: pt
pt simultaneous solution yields: t2
400k i.e.: t2
Answer: (1) When the piston has no pressure on the ground, the gas pressure is.
pa；(2) When the cylinder is just facing the ground without pressure, the temperature of the gas is 127
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Anonymous users2024-02-01According to the known conditions, the force analysis of the piston and cylinder was carried out respectively, and the balance equation was listed and calculated in combination with Charlie's law.
When the temperature t1"273+27=300"k, the piston has no pressure on the ground, and the column equilibrium equation:
p1s+mg=p0s to obtain p1
p0<>
pa-<>
Pa=Pa If the temperature increases, the gas pressure increases, and the cylinder is just facing the ground without pressure, the column equilibrium equation:
p2s=mg+p0s to obtain p2
p0<>
pa+<>
Pa = Pa According to Charlie's Law:
<>t="127" ℃
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Anonymous users2024-01-31<>2) q= <>
p0v+αt
1) In the gas by p=
During the descent to p0, the volume of the gas does not change, and the temperature is changed by t=""t0 becomes t1 and is <> by Charlie's law
The temperature of the gas is determined by t1
When the change to t0 changes, the volume decreases from v to v1
The gas pressure does not change, which is <> by the Gai-Lussac law
The solution is <>
2) In the process of piston descending, the work done by the piston on the gas is w=p0v-v1, in this process, the reduction of the internal energy of the gas is δu= (t1t0 is obtained by the first law of thermodynamics, the heat emitted by the gas in the cylinder is q=w+δu, and the solution q= <>
p0v+αt0