What pressure is exerted by 0.625 mol of a gas in a 45.4 L container at -24.0ºC
My chemistry is rusty, but since this is urgent, try this...
pV=nRT
45.4 L (p)= 0.625 (R) (-24 +273)
R= gas constant
T has to be changed into degrees Kelvin
then solve for p
I think it's right
Well, if I had to choose a pressure that 0.625 mol of gas would exert in a 45.4 L container at -24.0ºC, I'd probably say it's enough pressure to make you take a deep breath and say, "Whoa, that's a lot of gas!" But, to be more specific, we can calculate the pressure using the ideal gas law equation, which is pV = nRT.
First, we need to convert the temperature from Celsius to Kelvin. Just add 273.15 to -24.0, and you'll get 249.15 K. Now, let's substitute the values we have into the equation:
p x 45.4 = 0.625 x 8.3145 x 249.15
After solving the equation, the pressure comes out to be approximately 248.73 kPa. So, we can say that the gas exerts a pressure of around 248.73 kilopascals in the 45.4-liter container at -24.0ºC. But hey, don't let the numbers pressure you!
To find the pressure exerted by the gas, we can use the ideal gas law equation, which states:
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 temperature from Celsius to Kelvin. The conversion formula is:
T (K) = T (°C) + 273.15
Substituting the values into the equation:
T (K) = -24.0°C + 273.15
T (K) = 249.15 K
Now, we can plug the values into the ideal gas law equation:
PV = nRT
P * 45.4 L = 0.625 mol * 0.0821 L·atm/mol·K * 249.15 K
P * 45.4 L = 10.2074 L·atm
P = 10.2074 L·atm / 45.4 L
P = 0.2247 atm
Therefore, the pressure exerted by 0.625 mol of gas in a 45.4 L container at -24.0ºC is approximately 0.2247 atm.
To find the pressure exerted by the gas, we can use the ideal gas law equation, which is represented as:
PV = nRT
Where:
P is the pressure (in units of force per unit area, such as Pascal or atm)
V is the volume of the gas (in liters)
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/K·mol at standard conditions)
T is the temperature of the gas (in Kelvin)
First, we need to convert the temperature from Celsius to Kelvin by adding 273.15 to -24.0ºC:
T = -24.0ºC + 273.15 = 249.15 K
Now, we can substitute the given values into the ideal gas law equation and solve for pressure (P):
P * 45.4 L = 0.625 mol * 0.0821 L·atm/K·mol * 249.15 K
P * 45.4 L = 12.773 atm·L
P = 12.773 atm·L / 45.4 L
P ≈ 0.281 atm
Therefore, the pressure exerted by 0.625 mol of gas in a 45.4 L container at -24.0ºC is approximately 0.281 atm.