A sample of Helium occupies 200 L at 30ºC under a pressure of 841 torr. What is the volume it occupies at standard temprature and pressure (STP)?
Well, it looks like Helium really needs its personal space. But don't worry, I'll help you with the calculations.
To find the volume of Helium at STP (Standard Temperature and Pressure), we'll have to convert both temperature and pressure.
At STP, the temperature is 273.15 K and the pressure is 1 atmosphere (or 760 torr).
Using the combined gas law equation, which states that (P1 x V1)/(T1) = (P2 x V2)/(T2), let's solve for V2 (the unknown volume at STP).
First, let's convert the initial temperature to Kelvin by adding 273.15 to 30ºC, giving us 303.15 K.
Now, we can set up the equation:
(841 torr x 200 L) / (303.15 K) = (760 torr x V2) / (273.15 K)
Cross-multiplying and simplifying, we find:
(841 torr x 200 L x 273.15 K) / (303.15 K) = 760 torr x V2
Simplifying a bit further, we get:
V2 = (841 torr x 200 L x 273.15 K) / (760 torr x 303.15 K)
Crunching the numbers, I got V2 ≈ 152.41 L.
So, Helium really shrinks down at STP and occupies roughly 152.41 liters. Maybe it's just trying to fit in with the other gases? Who knows!
To find the volume of helium at standard temperature and pressure (STP), we need to use the ideal gas law equation, which states:
PV = nRT
where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature of the gas in Kelvin
At STP, the pressure is 1 atmosphere (atm) and the temperature is 273.15 Kelvin (K).
Given:
Initial volume (V1) = 200 L
Initial temperature (T1) = 30 ºC = 30 + 273.15 K
Initial pressure (P1) = 841 torr
First, we need to convert the temperature and pressure to the appropriate units:
P1 = 841 torr converted to atm => P1 = 841 torr / 760 torr/atm = 1.1066 atm
T1 = (30 + 273.15) K = 303.15 K
Now we can calculate the number of moles of helium using the ideal gas law:
PV = nRT
n = (PV) / (RT)
n = (1.1066 atm * 200 L) / (0.0821 L·atm/mol·K * 303.15 K)
n ≈ 9.549 moles
Next, we can use the number of moles and the new pressure and temperature at STP to find the new volume (V2):
P2 = 1 atm
T2 = 273.15 K
Using the ideal gas law equation again, we can solve for V2:
V2 = (n * R * T2) / P2
V2 = (9.549 moles * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm
V2 ≈ 215.6 L
Therefore, the volume of helium at standard temperature and pressure (STP) is approximately 215.6 L.
To determine the volume of helium at standard temperature and pressure (STP), we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
First, we need to convert the given temperature from 30ºC to Kelvin:
T(K) = T(ºC) + 273.15
T(K) = 30ºC + 273.15
T(K) = 303.15 K
Next, we can substitute the values into the equation and solve for the number of moles:
n = (PV) / (RT)
n = (841 torr * 200 L) / (0.0821 L·atm/mol·K * 303.15 K)
Now, let's solve for n:
n ≈ 69.52084 mol
Now, we can calculate the new volume (V2) at STP when the pressure (P2) is 1 atmosphere:
P1 * V1 = P2 * V2
(841 torr * 200 L) = (1 atm * V2)
V2 = (841 torr * 200 L) / (1 atm)
V2 = 168,200 torr·L / 1 atm
Since 1 atm is equal to 760 torr, we can simplify further:
V2 = 168,200 torr·L / 760 torr/atm
V2 ≈ 221.58 L
Therefore, the volume of helium at standard temperature and pressure (STP) is approximately 221.58 L.
Use
(P1V1/T1)=(P2V2/T2)
Standard conditions are P2 = 760 torr, T = 273 K. Don't forget to change 30 C to K.
K = 273 + C