a sample of helium gas has a volume of 6.50L at a pressure of 845mm Hg and a temperature of 25 celcius.what is the pressure of the gas in atm when the volume and temperature of the gas sample are changed to the following:
a. 1850 mL and 325K
To solve this problem, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, let's convert the given values to the appropriate units:
- The volume is given as 6.50 L, and we're asked to find the pressure in atm, so we don't need to convert anything.
- The initial pressure is given as 845 mm Hg. To convert this to atm, we need to divide by 760 (since 760 mm Hg = 1 atm). So, the initial pressure is 845/760 = 1.11 atm.
- The initial temperature is given as 25 °C. To convert this to Kelvin, we need to add 273.15 to the Celsius value. So, the initial temperature is 25 + 273.15 = 298.15 K.
Now let's solve the problem for the new conditions:
a. The new volume is given as 1850 mL. To convert this to liters, we divide by 1000. So, the new volume is 1850/1000 = 1.85 L.
The new temperature is given as 325 K.
Using the ideal gas law equation, we can set up the following equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Substituting the given and calculated values:
(1.11 atm * 6.50 L) / (298.15 K) = (P2 * 1.85 L) / (325 K)
Now, solve for P2:
P2 = [(1.11 atm * 6.50 L) / (298.15 K)] * (325 K / 1.85 L)
P2 = 3.29 atm
Therefore, the pressure of the helium gas in atm when the volume and temperature are changed to 1850 mL and 325 K, respectively, is 3.29 atm.