A sealed glass bottle containing air at atmospheric pressure (101kPa) and having a volume of 22 cm^3 is at 23 degrees Celsius. It is then tossed into an open fire.

When the temperature of the air in the bottle reaches 391 degrees celsius, what is the pressure inside the bottle? Assume any volume changes of the bottle are negligible. Answer in units of kPa.

Use the ideal gas law PV=nRT

V, n, and R are constant and irrelevant in this situation.

Therefore, use the equation P1/T1 = P2/T2.

101/23+273 = P2/391+273 (adding 273 because you MUST convert to Kelvin)

Solve for P2

Well, I must say, that bottle is in for a hot time! Now, let me calculate the pressure for you.

To solve this, we can use 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 in Kelvin.

First, we need to convert the initial and final temperatures from Celsius to Kelvin. So, 23 degrees Celsius + 273.15 = 296.15 K initial temperature, and 391 degrees Celsius + 273.15 = 664.15 K final temperature.

Since the bottle is sealed, the number of moles of air remains constant. Therefore, we can rewrite the equation as P₁V₁/T₁ = P₂V₂/T₂.

Now, let's plug in the values we have: P₁ = 101 kPa (initial pressure), V₁ = 22 cm³ (initial volume), T₁ = 296.15 K (initial temperature), T₂ = 664.15 K (final temperature).

101 kPa * 22 cm³ / 296.15 K = P₂ * 22 cm³ / 664.15 K

Simplifying this equation, we find that P₂ = (101 kPa * 22 cm³ * 664.15 K) / (296.15 K * 22 cm³).

Calculating that out, P₂ ≈ 226.87 kPa.

So, when the air in the sealed bottle reaches 391 degrees Celsius, the pressure inside will be approximately 226.87 kPa. Just remember, don't try this at home! Leave it to the professionals and avoid turning your friendly bottle into a hot air balloon.

To solve this problem, we can use the ideal gas law, which states:

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.

First, let's convert the initial temperature from Celsius to Kelvin:
T1 = 23 degrees Celsius + 273.15 = 296.15 K

Since the volume of the bottle is negligible and the number of moles of air remains constant, we can rewrite the ideal gas law as:

P1/T1 = P2/T2

where P1 is the initial pressure (101 kPa), and T2 is the final temperature (391 degrees Celsius + 273.15 = 664.15 K).

Now we can solve for P2:

P2 = (P1 * T2) / T1
P2 = (101 kPa * 664.15 K) / 296.15 K
P2 = 225.54 kPa

Therefore, when the temperature of the air in the bottle reaches 391 degrees Celsius, the pressure inside the bottle is approximately 225.54 kPa.

To solve this problem, we can use the ideal gas law equation, which relates the pressure, volume, temperature, and number of moles of a gas:

PV = nRT

Where:
P = Pressure
V = Volume
n = Number of moles
R = Ideal gas constant (8.31 J/(mol·K))
T = Temperature

First, let's convert the volume from cm^3 to m^3:
V = 22 cm^3 = 22/100^3 m^3 = 0.000022 m^3

We are given that the initial pressure is 101 kPa, the final temperature is 391 degrees Celsius, and the final pressure is what we need to find.

Next, we need to convert the temperatures from Celsius to Kelvin:
T_initial = 23 degrees Celsius + 273.15 = 296.15 K
T_final = 391 degrees Celsius + 273.15 = 664.15 K

Since the volume change of the bottle is negligible, we can assume that the number of moles of air in the bottle remains constant throughout the process.

Now, we can rearrange the ideal gas law equation to solve for the final pressure (P_final):
P_final = (n * R * T_final) / V

Since the number of moles (n), the ideal gas constant (R), and the volume (V) are constants:

P_final = (n * R * T_final) / V_initial

Substituting the given values:
P_final = (101 kPa * 0.000022 m^3) / (296.15 K * 8.31 J/(mol·K))

Performing the calculations:
P_final = 0.0002222 kPa

Therefore, when the temperature of the air in the bottle reaches 391 degrees Celsius, the pressure inside the bottle will be approximately 0.0002222 kPa.