Can you please tell me if these are right?
A gas has a volume of 2.8 L at a temperature of 27 degrees C. What temperature degrees C is needed to expand the volume to 15 L? (P and n are constant.)
I got 1334 degress C
Combined gas law problem a balloon is filled with 500.0 mL if helium at a temperature of 27 degress C and 755 mm Hg. As the baloon rises in the atmoshper the pressure and temperature drop what volume will it have when it reaches an altiude where the temperature is -33 degrees C and the prssure is 0.65 atm.
I got 609 mL.
Thanks.
I agree with #1. #2 I get 611 mL.
To solve these problems, we can use the ideal gas law equation 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.
Let's analyze the first problem:
Given:
Volume of gas (V1) = 2.8 L
Initial temperature (T1) = 27 degrees C = 300.15 K
Final volume (V2) = 15 L
We need to find the final temperature (T2).
Step 1: Convert the initial and final temperatures to Kelvin:
T1 = 27 + 273.15 = 300.15 K
Step 2: Use the formula:
(V1/T1) = (V2/T2)
Step 3: Rearrange the formula to solve for T2:
T2 = (V2 * T1) / V1
Substituting the given values:
T2 = (15 * 300.15) / 2.8
T2 ≈ 1607.68 K
So, the temperature required to expand the volume to 15 L is approximately 1607.68 degrees Celsius.
Now let's move on to the second problem:
Given:
Initial volume (V1) = 500.0 mL = 0.5 L
Initial temperature (T1) = 27 degrees C = 300.15 K
Initial pressure (P1) = 755 mm Hg
Final temperature (T2) = -33 degrees C = 240.15 K
Final pressure (P2) = 0.65 atm
We need to find the final volume (V2).
Step 1: Convert the initial volume and pressure to liters and atmospheres, respectively:
V1 = 0.5 L
P1 = 755 mm Hg * (1 atm / 760 mm Hg) ≈ 0.9934 atm
Step 2: Use the formula:
(P1 * V1)/ T1 = (P2 * V2) / T2
Step 3: Rearrange the formula to solve for V2:
V2 = (P1 * V1 * T2) / (P2 * T1)
Substituting the given values:
V2 = (0.9934 * 0.5 * 240.15) / (0.65 * 300.15)
V2 ≈ 0.609 L
So, the volume of the balloon when it reaches the given altitude is approximately 0.609 L or 609 mL.
These calculations should give you the correct answers for both problems.
To solve these problems, you would need to use the combined gas law equation, which relates the pressure, volume, and temperature of a gas.
The combined gas law equation is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
where P1, V1, and T1 are the initial pressure, volume, and temperature of the gas, and P2, V2, and T2 are the desired pressure, volume, and temperature.
Let's solve the first problem:
Given:
Initial volume (V1) = 2.8 L
Initial temperature (T1) = 27 degrees C
Final volume (V2) = 15 L
We are asked to find the final temperature (T2).
Rearranging the equation, we get:
T2 = (P1 * V1 * T2) / (P2 * V2)
Plugging in the given values:
P1 and P2 are constant, so we can ignore them.
T2 = (2.8 L * (27 + 273.15) K) / (15 L)
Calculating this expression, we get:
T2 ≈ 502.11 K
Converting back to degrees Celsius, we subtract 273.15 from the result:
T2 ≈ 502.11 K - 273.15 ≈ 228.96 degrees C
Therefore, the temperature needed to expand the volume to 15 L is approximately 228.96 degrees Celsius.
Now let's solve the second problem:
Given:
Initial volume (V1) = 500.0 mL = 0.5 L
Initial temperature (T1) = 27 degrees C = 27 + 273.15 K = 300.15 K
Initial pressure (P1) = 755 mm Hg
Final temperature (T2) = -33 degrees C = -33 + 273.15 K = 240.15 K
Final pressure (P2) = 0.65 atm
We are asked to find the final volume (V2).
Rearranging the equation, we get:
V2 = (P1 * V1 * T2) / (P2 * T1)
Plugging in the given values:
V2 = (755 mm Hg * 0.5 L * 240.15 K) / (0.65 atm * 300.15 K)
Before calculating, let's convert mm Hg to atm by dividing by 760:
V2 = (0.99342 atm * 0.5 L * 240.15 K) / (0.65 atm * 300.15 K)
Calculating this expression, we get:
V2 ≈ 0.609 L
Therefore, when the balloon reaches an altitude where the temperature is -33 degrees C and the pressure is 0.65 atm, it will have a volume of approximately 0.609 liters or 609 mL.
So to summarize:
For the first problem, the correct answer is approximately 228.96 degrees Celsius.
For the second problem, the correct answer is approximately 0.609 liters or 609 mL.