5.67 mol of N20 fills a ball that has a volume of 0.45 L and exerts 5.6 atm at 78K. What is the temperature when the volume is increased to 1.98L? Your answer must have the correct units
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
From the initial conditions:
P1 = 5.6 atm
V1 = 0.45 L
n = 5.67 mol
T1 = 78 K
Solving for the initial temperature T1:
T1 = (P1 * V1) / (n * R)
T1 = (5.6 * 0.45) / (5.67 * 0.0821)
T1 = 46.67 K
Now, we can find the final temperature T2 when the volume is increased to 1.98 L:
P2 = P1
V2 = 1.98 L
T2 = (P2 * V2) / (n * R)
T2 = (5.6 * 1.98) / (5.67 * 0.0821)
T2 = 77.03 K
Therefore, the temperature when the volume is increased to 1.98 L is 77.03 K.