A sample of argon gas occupies a volume of 1.99 L at -44.0 oC at 1.19 atm. What would its Celsius temperature be at 2.06 atm if its volume decreases to 1.53 L?
To solve this problem, we can use the combined gas law equation:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = 1.19 atm
V1 = 1.99 L
T1 = -44.0 oC + 273.15 = 229.15 K
P2 = 2.06 atm
V2 = 1.53 L
T2 = ?
Plugging in the values, we get:
(1.19 * 1.99) / 229.15 = (2.06 * 1.53) / T2
Solving for T2:
T2 = (2.06 * 1.53 * 229.15) / (1.19 * 1.99)
T2 = 359.29 K
Converting T2 back to Celsius:
T2 = 359.29 - 273.15
T2 = 86.14 oC
Therefore, the Celsius temperature of the argon gas would be 86.14 oC at 2.06 atm and 1.53 L.