Question

A sample of 2.00 mol of gas in a 10.00 L container is at 45.0 °C. What is the pressure (in atm) of the gas?

Answer 1

Answer:

5.22 atm

Explanation:

The following data were obtained from the question:

Number of mole (n) = 2 moles

Volume (V) = 10 L

Temperature (T) = 45 °C

Pressure (P) =?

Next, we shall convert 45 °C to Kelvin temperature. This can be obtained as follow:

Temperature (K) = Temperature (°C) + 273

T (K) = T (°C) + 273

T (°C) = 45 °C

T(K) = 45 °C + 273

T (K) = 318 K

Finally, we shall determine the pressure of the gas by using the ideal gas equation as shown below:

Number of mole (n) = 2 moles

Volume (V) = 10 L

Temperature (T) = 318 K

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

Pressure (P) =.?

PV = nRT

P x 10 = 2 x 0.0821 x 318

Divide both side by 10

P = (2 x 0.0821 x 318) /10

P = 5.22 atm

Therefore, the pressure of the gas is 5.22 atm