sealed metal container can withstand a pressure difference of 0.630 . The container initially is filled with an ideal gas at 12 and 0.80 To what temperature can you cool the container before it collapses? (Ignore any changes in the container's volume due to thermal expansion
Numbers without dimensions are meaningless. Is the 0.630 atmospheres? Is the 12 degrees C? What is the 0.80?
We are not mindreaders here.
To solve this problem, we can apply the ideal gas law, which states that 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.
First, we need to find the number of moles of gas in the container. We can use the ideal gas law again by rearranging it as n = PV / RT.
Given:
Pressure difference (ΔP) = 0.630 atm
Initial pressure (P) = 12 atm
Volume (V) = 0.80 L
Ideal gas constant (R) = 0.0821 L·atm/(mol·K)
Plugging in the values, we can calculate the number of moles:
n = (12 atm * 0.80 L) / (0.0821 L·atm/(mol·K) * T)
Next, we need to determine the maximum pressure the container can withstand before collapsing. This pressure is the sum of the initial pressure and the pressure difference, so:
Maximum pressure (P_max) = P + ΔP
Finally, we can solve for the temperature (T) by rearranging the ideal gas law:
T = (P_max * V) / (n * R)
Plugging in the values, we can calculate the temperature at which the container will collapse.