L has twice the volume of container J. Container J is at a temperature of 100 K, and container L is at 200 K. How does the pressure in container J compare with that in container L
It depends on what is inside the container. If a gas
Then the higher temperature is twice the pressure of the lower temperature container.
P1/P2=T1/T2 Read about Charles' Law http://en.wikipedia.org/wiki/Charles%27s_law
To compare the pressure in container J with container L, we need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
Given that container J has a volume of Vj, and container L has twice the volume of container J, then container L would have a volume of Vl = 2Vj.
We are told that container J is at a temperature of 100 K, and container L is at 200 K.
Let's assume that the number of moles of gas in both containers is the same, so nJ = nL = n.
Now, let's compare the pressures in the two containers using the ideal gas law equation.
For container J:
PJ * Vj = n * R * TJ
For container L:
PL * Vl = n * R * TL
Since we are comparing the pressures, let's divide the two equations:
(PJ * Vj) / (PL * Vl) = (n * R * TJ) / (n * R * TL)
Simplifying:
Vj / Vl = TJ / TL
Since Vl is twice the volume of Vj, we can substitute Vl = 2Vj:
Vj / (2Vj) = 100 K / 200 K
1/2 = 1/2
Therefore, the pressure in container J is equal to the pressure in container L.
In conclusion, the pressure in container J is the same as the pressure in container L.