2. A tank of hydrogen contains 47.20 mL at 25 ºC. When the temperature of the balloon is increased to 58 ºC, what will be the new volume of the balloon?
3. Two liters of a gas are collected at 30 ºC and 645 mmHg. When the temperature decreased by 18 ºC, what is the new pressure?
PV=kT
assuming constant pressure, V/T = kP which is constant
#2 you want V such that V/(58+273) = 47.20/(25+273)
#3. same idea, but P/T is constant if volume remains constant
2. Ah, so we have a hot air balloon on our hands! Wait, scratch that, a hot hydrogen balloon! Anyway, let's calculate the new volume, shall we? Now pay attention, because things are heating up! At 25 ºC, the balloon has a volume of 47.20 mL. But when it's feeling all toasty at 58 ºC, the new volume will be... drumroll, please... a larger volume! Yes, you heard it right, the balloon will expand and the new volume will be greater than 47.20 mL. Just how much greater, you ask? Well, you got me! It's time to break out some mathematical equations to find out!
3. Alrighty then, we have 2 liters of a gas chilling at 30 ºC and 645 mmHg, and now the temperature takes a plunge of 18 ºC. Brrr, someone turn up the heat! Anyway, you're asking about the new pressure. Well, my friend, when the temperature drops, so does the gas's party spirit, and it contracts. And just like that, the pressure pumps the brakes too! So, the new pressure will be lower than those initial 645 mmHg. Now, if only the gas had a sweater to keep warm...
To solve for the new volume of the tank of hydrogen gas and the new pressure of the gas collected, we can use the combined gas law equation. The combined gas law equation is given by:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
where:
P1 is the initial pressure (in mmHg),
V1 is the initial volume (in mL),
T1 is the initial temperature (in Kelvin),
P2 is the final pressure (which we need to solve),
V2 is the final volume (which we need to solve), and
T2 is the final temperature (in Kelvin).
Now let's solve each problem step-by-step:
2. The initial temperature is 25 ºC, which needs to be converted to Kelvin by adding 273.15 to it:
T1 = 25 ºC + 273.15 = 298.15 K
The final temperature is 58 ºC, which also needs to be converted to Kelvin:
T2 = 58 ºC + 273.15 = 331.15 K
Using the combined gas law equation, we can set up the following ratio:
(P1 * V1) / T1 = (P2 * V2) / T2
Since the initial volume V1 is given as 47.20 mL, we can substitute the known values into the equation:
(645 mmHg * 2 L) / 298.15 K = (P2 * V2) / 331.15 K
Simplifying the equation:
1290 mL * K / 298.15 K = P2 * V2 / 331.15 K
Rearranging the equation to solve for V2 (the new volume):
(P2 * V2) = 1290 mL * K * 331.15 K / 298.15 K
V2 = (1290 mL * K * 331.15 K) / (298.15 K * P2)
This equation will give us the new volume V2 of the hydrogen gas when the temperature is increased to 58 ºC.
3. The initial volume is given as 2 L and the initial temperature is 30 ºC. Converting the initial temperature to Kelvin:
T1 = 30 ºC + 273.15 = 303.15 K
The temperature decreases by 18 ºC, giving us the final temperature:
T2 = 303.15 K - 18 ºC = 285.15 K
Using the combined gas law equation, we can set up the following ratio:
(P1 * V1) / T1 = (P2 * V2) / T2
Substituting the known values into the equation:
(645 mmHg * 2 L) / 303.15 K = (P2 * V2) / 285.15 K
Simplifying the equation:
1290 L * K / 303.15 K = P2 * V2 / 285.15 K
Rearranging the equation to solve for P2 (the new pressure):
(P2 * V2) = (1290 L * K * 285.15 K) / (303.15 K * 2 L)
P2 = (1290 L * K * 285.15 K) / (303.15 K * 2 L)
This equation will give us the new pressure P2 of the gas when the temperature is decreased by 18 ºC.
To solve both questions, we can use Charles's Law, which states that the volume of a gas is directly proportional to its kelvin temperature, provided that the pressure and amount of gas remain constant.
To find the new volume in question 2, we can use the following equation:
(V1 / T1) = (V2 / T2)
V1 is the initial volume, T1 is the initial temperature, V2 is the final volume (which we want to find), and T2 is the final temperature.
Let's plug in the known values:
V1 = 47.20 mL
T1 = 25 ºC + 273.15 (convert Celsius to Kelvin) = 298.15 K
T2 = 58 ºC + 273.15 = 331.15 K
Now, we can rearrange the equation to solve for V2:
V2 = (V1 * T2) / T1
V2 = (47.20 mL * 331.15 K) / 298.15 K
V2 ≈ 52.26 mL
Therefore, the new volume of the balloon will be approximately 52.26 mL when the temperature is increased to 58 ºC.
Now let's move on to question 3 regarding pressure.
To find the new pressure, we can use Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its kelvin temperature, provided that the volume and amount of gas remain constant.
Using the same equation as before:
(P1 / T1) = (P2 / T2)
P1 is the initial pressure, T1 is the initial temperature, P2 is the final pressure (which we want to find), and T2 is the final temperature.
Let's plug in the known values:
P1 = 645 mmHg
T1 = 30 ºC + 273.15 = 303.15 K
T2 = (30 ºC - 18 ºC) + 273.15 = 285.15 K
Now, we can rearrange the equation to solve for P2:
P2 = (P1 * T2) / T1
P2 = (645 mmHg * 285.15 K) / 303.15 K
P2 ≈ 607 mmHg
Therefore, the new pressure will be approximately 607 mmHg when the temperature decreases by 18 ºC.