A steel tank at 300 K contains 0.285 L of gas at 1.92 atm pressure. The tank is capable of withstanding a maximum pressure of 5.76 atm. Assuming that doubling the kelvin temperature causes the internal pressure to double,

a. at what temperature will the tank burst?

b. will it burst if placed in a fire burning at 1275 K?

I was told to use (V1/T1) = (V2/T2) but do I keep the same volume?

I misspoke. I should have said (P1/T1) = (P2/T2)

a. You can work a without any work except reasoning on your part. If you double the pressure that will be 1.92*2 = 3.84 atm and that is less than 5.76 so that tank should not rupture.

b.
Use the above p/t formula.
P1 = 1.92 atm
T1 = 300 K
P2 = ?
T2 = 1275 K.

(1.92/300) = (P2/1275)
Solve for P2 and compare with the rupture pressure of 5.76 atm.

Ah, the tank is holding its breath. Let's see if we can help it out with some calculations, while adding a dash of humor, of course!

According to the given information, we know that doubling the Kelvin temperature causes the internal pressure to double as well. And it's always important to keep the volume constant when using gas laws.

a. So, if we assume the volume remains constant (0.285 liters in this case), we can use the formula (V1/T1) = (V2/T2) to find the temperature at which the tank bursts.

Since the initial temperature (T1) is 300 K, and the initial pressure (P1) is 1.92 atm, we need to find the temperature (T2) at which the pressure (P2) reaches the maximum pressure the tank can withstand, which is 5.76 atm.

(0.285/300) = (0.285/T2)
T2 = (0.285/300) * (5.76/1.92)

Calculating this, we find that T2 is approximately 1.15 K.

So, the tank would burst at around 1.15 K. Brr, that's cold!

b. Now, let's warm things up a bit by considering the fire burning at 1275 K. Will the tank burst?

Well, since the tank can withstand a maximum pressure of 5.76 atm, and the fire doesn't exceed that pressure, the tank should be safe from bursting. It might feel a bit toasty, but at least it won't burst into flames!

Remember, it's always important to handle volatile situations with care and a touch of humor. Stay safe, my friend!

Yes, you would keep the same volume in this case since the problem does not mention any change in volume. The formula you mentioned, (V1/T1) = (V2/T2), is known as the combined gas law and is used to relate the initial and final pressures and temperatures of a gas under the assumption that the volume remains constant.

To answer your questions step-by-step:

a. To determine the temperature at which the tank will burst, we can set up the equation using the combined gas law:

(P1/T1) = (P2/T2)

Given:
P1 = 1.92 atm
T1 = 300 K
P2 = 5.76 atm (maximum pressure the tank can withstand)

Rearranging the equation to solve for T2:

T2 = (P2/T1) * T1
T2 = (5.76 atm / 1.92 atm) * 300 K
T2 = 3 * 300 K
T2 = 900 K

So, the tank will burst at a temperature of 900 K.

b. Now, to determine if the tank will burst in a fire burning at 1275 K, we need to compare the temperature of the fire (T3) with the bursting temperature of the tank (T2).

If doubling the Kelvin temperature causes the internal pressure to double, we can assume that the same relationship applies to the bursting temperature.

Let's compare T3 with T2:

T3 = 1275 K
T2 = 900 K

Since T3 is greater than T2, the temperature of the fire burning at 1275 K is higher than the temperature at which the tank will burst. Therefore, the tank will burst if placed in a fire burning at 1275 K.

To solve this question, you can use the ideal gas law: 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.

a. To find the temperature at which the tank will burst, we need to determine the pressure at which the tank will reach its maximum capacity. We are given that the tank can withstand a maximum pressure of 5.76 atm, and that doubling the Kelvin temperature causes the internal pressure to double.

Let's first find the current pressure using the ideal gas law. Given:
- Volume (V1) = 0.285 L
- Pressure (P1) = 1.92 atm
- Temperature (T1) = 300 K

Using the ideal gas law equation, we can rearrange it to solve for the unknown pressure (P2):
P2 = (P1 x T2) / T1

Since we want to find the temperature at which the tank will burst, we can assume that P2 is equal to the maximum pressure the tank can withstand, which is 5.76 atm. Thus, we can set up the equation:

5.76 atm = (1.92 atm x T2) / 300 K

Now, let's solve for T2:

T2 = (5.76 atm x 300 K) / 1.92 atm
T2 ≈ 900 K

Therefore, the temperature at which the tank will burst is approximately 900 K.

b. We need to determine if the tank will burst if it is placed in a fire burning at 1275 K. Given that doubling the Kelvin temperature causes the internal pressure to double, we can use this information to compare the pressure inside the tank at room temperature (300 K) with the pressure inside the tank at the fire temperature (1275 K).

Again, using the ideal gas law, we can set up the equation:

P1/P2 = T1/T2

where:
- P1 is the pressure at room temperature
- T1 is the room temperature
- P2 is the pressure at the fire temperature
- T2 is the fire temperature

We know that doubling the Kelvin temperature causes the internal pressure to double, so we can use this relationship:

P1/P2 = T1/T2 = 1/2

Since we know P1 is 1.92 atm, we can solve for P2:

1.92 atm / P2 = 1/2

P2 = 3.84 atm

As we can see, the pressure inside the tank at the fire temperature (3.84 atm) is less than the maximum pressure the tank can withstand (5.76 atm). Therefore, the tank will not burst if placed in a fire burning at 1275 K.