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 in Kelvin.
Let's start by converting the given temperatures to Kelvin: T1 = 27°C + 273.15 = 300.15 K (initial temperature) T2 = ? (final temperature)
Now, let's calculate the initial number of moles (n1) using the ideal gas law: P1V1 = n1RT1 (1.2 atm) * (5 L) = n1 * (0.0821 L·atm/(mol·K)) * (300.15 K) 6 = n1 * 24.980115 n1 = 6 / 24.980115 n1 ≈ 0.2402 moles
Since the number of moles of gas remains constant during the transfer, we can set n1 = n2 (n2 is the number of moles in the final container).
Now, let's calculate the final temperature (T2) using the ideal gas law again: P2V2 = n2RT2 (1 atm) * (3 L) = n1 * (0.0821 L·atm/(mol·K)) * T2 3 = 0.2402 * 0.0821 * T2 T2 ≈ 3 / (0.2402 * 0.0821) T2 ≈ 152.3 K
Finally, let's convert the final temperature from Kelvin to Celsius: T2 = 152.3 K - 273.15 ≈ -120.85°C
Therefore, the gas must be stored at approximately -120.85°C to achieve the desired conditions.