1) A container of volume V1= 2 liters is filled with an ideal gas under the pressure P1 =0.8 Bar at the temperature T1 =300K. 1)Calculate the quantity of matter present in this container. 2) The previous container noted (1) is connected by a tube of negligible volume to a container (2) of volume V2 =V1 and initially containing the same gas at the same temperature and pressure. The container (2) is heated to the temperature T2 =360K. We neglect the thermal exchanges between the two containers via the tube. Calculate the quantities of matter and the pressure in each container at mechanical equilibrium.
Question
- A container of volume V1= 2 liters is filled with an ideal gas under the pressure P1 =0.8 Bar at the temperature T1 =300K. 1)Calculate the quantity of matter present in this container.
- The previous container noted (1) is connected by a tube of negligible volume to a container (2) of volume V2 =V1 and initially containing the same gas at the same temperature and pressure. The container (2) is heated to the temperature T2 =360K. We neglect the thermal exchanges between the two containers via the tube. Calculate the quantities of matter and the pressure in each container at mechanical equilibrium.
Solution
To solve this problem, we will use the ideal gas law, which states that the pressure of a gas times its volume is equal to the number of moles of the gas times the gas constant times the temperature of the gas (PV = nRT).
- To calculate the quantity of matter (n1) in the first container, we rearrange the ideal gas law to solve for n:
n1 = P1V1 / RT1
Where: P1 = 0.8 Bar = 0.8 x 10^5 Pa (since 1 Bar = 10^5 Pa) V1 = 2 liters = 2 x 10^-3 m^3 (since 1 liter = 10^-3 m^3) R = 8.314 J/(mol.K) (universal gas constant) T1 = 300K
Substituting these values in, we get:
n1 = (0.8 x 10^5 Pa x 2 x 10^-3 m^3) / (8.314 J/(mol.K) x 300K) = 0.064 mol
- When the two containers are connected and the second container is heated, the total volume, total number of moles, and total temperature can be used to find the final pressure (Pf) and the number of moles in each container (n1f and n2f).
The total volume Vt = V1 + V2 = 2V1 (since V2 = V1) The total number of moles nt = 2n1 (since the same gas was initially in both containers) The total temperature Tt = (T1 + T2) / 2 = (300K + 360K) / 2 = 330K (since the containers are at mechanical equilibrium)
We can find the final pressure using the ideal gas law:
Pf = ntRTt / Vt
Substituting the values in, we get:
Pf = (2 x 0.064 mol x 8.314 J/(mol.K) x 330K) / (2 x 2 x 10^-3 m^3) = 0.87 Bar
The number of moles in each container at mechanical equilibrium is equal since the volumes are equal, so:
n1f = n2f = nt / 2 = 0.064 mol
So, the quantity of matter in each container is 0.064 mol and the pressure in each container is 0.87 Bar.
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