Question

0.24 moles of a gaseous product from a reaction is collected in a 25.0-L container at a pressure of 300.0 kPa. Determine the temperature (in °C) at which the gas was collected.

Answer 1

3428.58 °C

Step-by-step explanation:

We'll begin by converting 300 KPa to atm. This can be obtained as follow:

101.325 KPa = 1 atm

Therefore,

300 KPa = 300 KPa × 1 atm / 101.325 KPa

300 KPa = 2.96 atm

Next, we shall determine the temperature. This can be obtained as follow:

Number of mole (n) = 0.24 mole

Volume (V) = 25 L

Pressure (P) = 2.96 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) =?

PV = nRT

2.96 × 25 = 0.24 × 0.0821 × T

74 = 0.24 × 0.0821 × T

Divide both side by 0.25 × 0.0821

T = 74 / (0.24 × 0.0821)

T = 3755.58 K

Finally, we shall convert 3755.58 K to °C. This can be obtained as followb:

T(°C) = T(K) – 273

T(K) = 3755.58 K

T(°C) = 3755.58 – 273

T(°C) = 3428.58 °C

Thus, the temperature at which the gas was collected is 3428.58 °C