What is the temperature (in degrees Celsius) of 6.02 mol of hydrogen sulfate gas in a 100 L container at 0.95 atm?
P = 0.95 atm
V = 100 L
n = 6.02 mol
R = 0.0821 atm L mol-1 K-1
T = unknown
The Ideal Gas equation can be used.
The Ideal Gas equation can be expressed as: PV = nRT
To find the temperature, solve the equation for T:
T = PV / nR
T = (0.95 atm × 100 L) / (6.02 mol × 0.0821 atm L mol-1 K-1) = 192.2 K
T (K) = t (°C) + 273.15
t = T - 273.15 = 192.2 - 273.15 = -80.95°C = -81°C
t = -81°C