If 5.00L of argon gas is at 0.460 atm and -123C what is the volume at STP
I think it's b
To determine the volume of argon gas at STP (Standard Temperature and Pressure), we need to use the ideal gas law equation, which is:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Given:
Initial pressure (P1) = 0.460 atm
Initial volume (V1) = 5.00 L
Initial temperature (T1) = -123°C = -396 K
STP conditions:
Pressure at STP (P2) = 1 atm
Temperature at STP (T2) = 273 K
To find the number of moles of argon gas (n), we can rearrange the ideal gas law equation:
n = (PV) / (RT)
Now let's plug in the values:
n1 = (P1 * V1) / (R * T1)
n1 = (0.460 atm * 5.00 L) / (R * -396 K)
Since R is the ideal gas constant and has a fixed value of 0.0821 L·atm/(K·mol), we can substitute it in:
n1 = (0.460 atm * 5.00 L) / (0.0821 L·atm/(K·mol) * -396 K)
Calculate n1 to find the number of moles.
Once you have n1, you can rearrange the ideal gas law equation to find the volume at STP (V2):
V2 = (n2 * R * T2) / P2
Substitute the values in:
V2 = (n1 * 0.0821 L·atm/(K·mol) * 273 K) / 1 atm
Calculate V2 to find the volume at STP.
To find the volume of argon gas at STP (Standard Temperature and Pressure), we can use the ideal gas law equation, which is:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles of gas
R = Ideal Gas Constant (0.0821 L·atm/mol·K)
T = Temperature (in Kelvin)
First, we need to convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = -123°C + 273.15
T(K) = 150.15 K
Now we can calculate the number of moles of argon gas using the ideal gas law:
PV = nRT
n = PV / RT
n = (0.460 atm)(5.00 L) / (0.0821 L·atm/mol·K)(150.15 K)
n ≈ 0.1372 mol
Since the volume of gas at STP is given at standard temperature and pressure, we can assume that the pressure is 1 atm, and the temperature is 0°C or 273.15 K.
Now we can rearrange the ideal gas law equation to find the volume at STP:
V2 = (nR * T2) / P2
V2 = (0.1372 mol)(0.0821 L·atm/mol·K)(273.15 K) / (1 atm)
V2 ≈ 3.619 L
Therefore, the volume of argon gas at STP is approximately 3.619 liters.