A 828 mL sample of NO2 gas is at STP. If the volume changes to 598 mL and the temperature changes to 12ºC, what will be the new pressure in atmospheres?
new pressure=_____atm
assuming the gas is ideal, we can use the formula,
P1*V1/T1 = P2*V2/T2
where
P1 = initial pressure
V1 = initial volume
T1 = initial temperature (in K)
P2 = final pressure
V2 = final volume
T2 = final temperature (in K)
we first convert the given temp to Kelvin units:
T2 = 12 + 273.15 = 285.15 K
*then note that at STP, temp is at 273.15 K and 1 atm. substituting,
P1*V1/T1 = P2*V2/T2
(1)(828) / 273.15 = (P2)(598) / 285.15
P2 = (828)(285.15) / (273.15)(598)
P2 = 1.45 atm
hope this helps~ :)
thank you very much it does help a lot but when im doing this problems on my oown they dnt make any sense ..
To find the new pressure in atmospheres, we need to use the ideal gas law equation, which is:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = number of moles
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature (in Kelvin)
First, let's convert the initial volume from milliliters (mL) to liters (L) since the rest of the equation uses volume in liters. We can do this by dividing the initial volume by 1000:
Initial volume = 828 mL ÷ 1000 = 0.828 L
The initial volume is 0.828 L.
Since the question states that the gas is at STP (Standard Temperature and Pressure), we know that the initial conditions are:
Temperature (T1) = 0ºC (273 K)
Pressure (P1) = 1 atm
Now, let's calculate the number of moles (n) using the ideal gas law:
n = (P1 * V1) / (R * T1)
= (1 atm * 0.828 L) / (0.0821 L·atm/mol·K * 273 K)
≈ 0.036 moles (rounded to three decimal places)
Now, let's convert the new temperature from Celsius to Kelvin:
New temperature = 12ºC + 273 = 285 K
Now, we can substitute the known values into the ideal gas law equation to find the new pressure (P2):
P2 = (n * R * T2) / V2
= (0.036 moles * 0.0821 L·atm/mol·K * 285 K) / 0.598 L
≈ 0.484 atm (rounded to three decimal places)
Therefore, the new pressure is approximately 0.484 atm.