Answer:
110 °C
Explanation:
The following data were obtained from the question:
Initial pressure (P1) = 0.538 atm
Initial temperature (T1) = 256 °C
Initial volume (V1) = 437 mL.
Final pressure (P2) = constant = 0.538 atm
Final volume (V2) = 316 mL
Final temperature (T2) =?
Next we shall convert 256 °C to Kelvin temperature. This Can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T1) = 256 °C
Initial temperature (T1) = 256 °C + 273 = 529 K
Next, we shall determine the new temperature of the gas as illustrated below:
Pressure = constant
Initial temperature (T1) = 529 K
Initial volume (V1) = 437 mL.
Final volume (V2) = 316 mL
Final temperature (T2) =?
Applying the Charles' law equation, we have:
V1 /T1 = V2 /T2
437/529 = 316/T2
Cross multiply
437 × T2 = 529 × 316
437 × T2 = 167164
Divide both side by 437
T2 = 167164 / 437
T2 = 382.5 ≈ 383 K
Finally, we shall convert 383 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
T(K) = 383 K
T(°C) = 383 – 273
T(°C) = 110 °C
Therefore, the new temperature of the gas is 110 °C
