how do you solve the ideal gas law when you are given the temperature in celcius instead of kelvin.
Heres an example question. Calculate the volume occupied by 3.25 moles of carbon dioxide at 25degrees celcius and 750mm Hg.
The use of the gas law requires you to use T in Kelvin. To convert celsius (note the correct spelling of celsius) to Kelvin,
K = 273.15 + C.
so all i do is add 273.15 to the temperature in celsius( sorry about the typo. i no i spelled chemistry wrong too)and then multiply/divide everything?
Yes, but don't forget that if R is in liter-atm units, then pressure must be in atmospheres. You convert mm Hg pressure to atm by dividing by 760. For example, 750 mm Hg to atm =
750 x (1 atm/760) = ??
To solve the ideal gas law equation, you need to use the given information and ensure that all the units are in the correct form. The ideal gas law equation is:
PV = nRT
Where:
- P is the pressure of the gas (in this case, 750 mm Hg)
- V is the volume of the gas (what we are trying to find)
- n is the number of moles of gas (in this case, 3.25 moles of carbon dioxide)
- R is the ideal gas constant (0.0821 L·atm/mol·K)
- T is the temperature of the gas in Kelvin (we need to convert Celsius to Kelvin)
The first thing you need to do is convert the temperature from Celsius to Kelvin. To convert Celsius to Kelvin, you add 273.15 to the Celsius value. So, in this case, 25 degrees Celsius would be:
T = 25 + 273.15 = 298.15 K
Now that we have the temperature in Kelvin, we can proceed with the calculation.
To solve for V, we rearrange the ideal gas law equation to solve for V:
V = (nRT) / P
Plug in the values you have:
V = (3.25 moles * 0.0821 L·atm/mol·K * 298.15 K) / 750 mm Hg
Before proceeding further, we need to convert the pressure from mm Hg to atm. Since 1 atm = 760 mm Hg, we divide the given pressure by 760:
P = 750 mm Hg / 760 mm Hg/atm = 0.987 atm
Now substitute this value for P in the equation:
V = (3.25 moles * 0.0821 L·atm/mol·K * 298.15 K) / 0.987 atm
After evaluating the expression, you will find the volume, V.