A metal tank with volume 3.10 L will burst if the pressure of the gas it contains exceeds 100 atm.
(a) If 11.0 mol of an ideal gas is put into the tank at a temperature of 23.00C, to what temperature can the gas be warmed before the tank ruptures? You can ignore the thermal expansion of the tank.
(b) Based on your answer to part (a), is it reasonable to ignore the thermal expansion of the tank? Explain?
Please focus on part (b). Thank you very much.
YEs, the effect of temperature on a gas is much greater than the effect on the metal tank.
To determine if it is reasonable to ignore the thermal expansion of the tank, we need to consider the change in volume due to temperature increase.
The ideal gas law can be used to relate the pressure, volume, temperature, and the number of moles of gas:
PV = nRT
Where:
P = pressure of the gas
V = volume of the tank
n = number of moles of gas
R = ideal gas constant
T = temperature of the gas in Kelvin
(a) To what temperature can the gas be warmed before the tank ruptures?
We can rearrange the ideal gas law equation to solve for the temperature (T):
T = PV / (nR)
Given:
P = 100 atm
V = 3.10 L
n = 11.0 mol
R = 0.0821 L·atm/(K·mol)
Plugging in the values, we have:
T = (100 atm) * (3.10 L) / (11.0 mol * 0.0821 L·atm/(K·mol))
Calculating this, we find:
T ≈ 352 K
Therefore, the gas can be warmed to approximately 352 Kelvin before the tank ruptures.
(b) Is it reasonable to ignore the thermal expansion of the tank? Explain.
In this problem, we were explicitly told to ignore the thermal expansion of the tank. This suggests that the tank's volume does not change significantly with temperature. If the tank's volume did change significantly, it would affect the pressure inside the tank and therefore the temperature at which it would burst.
However, it is important to note that in real-world scenarios, the thermal expansion of the tank can be a significant factor, especially if the temperature change is large. Ignoring the thermal expansion of the tank is a simplifying assumption for this problem, but it may not be valid in all cases.
To answer part (b) of the question, we need to consider the effect of thermal expansion on the tank.
Thermal expansion refers to the increase in size or volume of a material as its temperature increases. In the case of the metal tank, if the gas inside it is heated, both the gas and the tank will expand.
If the temperature is increased, the gas particles will move faster and collide more frequently with the tank walls. As a result, the gas will exert a higher pressure on the walls, potentially leading to the tank bursting.
When we calculate the maximum temperature the gas can be warmed to before the tank ruptures, we assume that the tank volume remains constant. However, in reality, the tank will also expand as it gets hotter.
In this case, we can examine the difference in thermal expansion between the gas and the tank. If the thermal expansion of the gas is significantly larger than that of the tank, then ignoring the thermal expansion of the tank may be reasonable.
To determine this, you would need to know the coefficient of thermal expansion for both the gas and the tank material. The coefficient of thermal expansion measures the change in size or volume of a material per unit change in temperature. Based on the values of the coefficients, you can compare the relative expansion of the gas and the tank.
If the gas expands much more than the tank, then neglecting the thermal expansion of the tank is reasonable. However, if the tank expands significantly compared to the gas, then the ideal gas law alone cannot accurately predict the behavior of the system, and the thermal expansion of the tank must be taken into account.
There isn't enough information provided in the original question to determine whether the thermal expansion of the tank can be ignored. To make a proper determination, you would need to know the coefficient of thermal expansion for both the gas and the tank material, as well as the specific temperatures involved.