A sample of gas has a volume of 1.30 L at STP. What is the temperature in °C if the volume is 13.0 L at 2.00 atm?
To determine the temperature in °C, we can use the combined gas law equation:
(P1V1) / (T1) = (P2V2) / (T2)
Where:
P1 = initial pressure (in this case, the pressure at STP)
V1 = initial volume (1.30 L)
T1 = initial temperature (at STP, 0 °C or 273.15 K)
P2 = final pressure (2.00 atm)
V2 = final volume (13.0 L)
T2 = final temperature (what we need to find)
Let's rearrange the equation to solve for T2:
T2 = (P2V2 * T1) / (P1V1)
Now we can substitute the given values into the equation:
T2 = (2.00 atm * 13.0 L * 273.15 K) / (1.00 atm * 1.30 L)
Now we can calculate T2:
T2 = (54.99 atm * L * K) / (atm * L)
T2 = 423.45 K
To convert from Kelvin to Celsius, subtract 273.15:
T2 = 423.45 K - 273.15 K
T2 = 150.30 °C
Therefore, the temperature in °C when the volume is 13.0 L at 2.00 atm is 150.30 °C.