Question

A sample of helium gas at a pressure of 0.538 atm and a temperature of 256 °C, occupies a volume of 437 mL. If the gas is cooled at constant pressure until its volume is 316 mL, the temperature of the gas sample will be

Answer 1

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