A plastic cylinder full of acetylene (C2H2) has a rigid volume of 2.92 L and a pressure of 1.80 atm at 20 °C. The cylinder can only withstand a pressure of 5 atm. Left inside a hot car where the temperature can reach 50 °C, is there danger of the cylinder rupturing? If so, at what temperature will the rupture take place?
There is just one problem here although can make two out of it. I would calculate the T necessary to achieve a pressure of 5 atm. Compare with T = 50 C.
Remember T must be in kelvin.
would i be using the ideal gas equation in order to calculate the T necessary?
To determine if there is danger of the cylinder rupturing, we need to calculate the new pressure inside the cylinder at the higher temperature and compare it to the maximum withstand pressure of the cylinder, which is 5 atm.
To calculate the new pressure, we can use the combined gas law equation, which relates the initial and final states of a gas sample:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 is the initial pressure,
V1 is the initial volume,
T1 is the initial temperature,
P2 is the final pressure (what we want to find),
V2 is the final volume (which remains the same since the cylinder is rigid), and
T2 is the final temperature.
We can rearrange the equation to solve for P2:
P2 = (P1 * V1 * T2) / (V2 * T1)
Given:
P1 = 1.80 atm,
V1 = 2.92 L,
T1 = 20 °C = 293 K,
V2 = 2.92 L (remains the same as the cylinder is rigid), and
T2 = 50 °C = 323 K.
Plugging the values into the equation, we get:
P2 = (1.80 atm * 2.92 L * 323 K) / (2.92 L * 293 K)
P2 = 4.95 atm
Upon calculating, we find that the new pressure inside the cylinder at 50 °C is approximately 4.95 atm.
Since the new pressure of 4.95 atm exceeds the maximum withstand pressure of the cylinder (5 atm), there is indeed danger of the cylinder rupturing if left inside a hot car where the temperature can reach 50 °C.
To find the temperature at which the rupture would occur, we can rearrange the equation to solve for T2:
T2 = (P2 * V2 * T1) / (P1 * V1)
Plugging in the values:
T2 = (5 atm * 2.92 L * 293 K) / (1.80 atm * 2.92 L)
Simplifying, we get:
T2 = 500.78 K
Therefore, the rupture would likely occur at a temperature of approximately 500.78 K.