A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mmHg. The flask is opened and more gas is added to the flask. The new pressure is 795 mmHg and the temperature is now 26 °C. There are now mol of gas in the flask.

The new moles of the gas in the flask is 2.13 moles.

Explanation:

Given;

number of moles of gas, n = 1.9 mol

temperature of the gas, T = 21 °C  = 21 + 273 = 294 K

pressure of gas, P = 697 mmHg

volume of gas, V = ?

Apply ideal gas law;

PV = nRT

Where;

R is gas constant, = 62.363 mmHg.L / mol. K

V = nRT / P

V = (1.9 x 62.363 x 294) / 697

V = 49.98 L

New pressure of the gas, P = 795 mmHg

New temperature of the gas, T = 26 °C = 273 + 26 = 299 K

New moles of the gas, n = ?

Volume of the gas is constant because volume of the flask is the same when more gas was added.

n = PV / RT

n = (795 x 49.98) / (62.363 x 299)

n = 2.13 moles

Therefore, the new moles of the gas in the flask is 2.13 moles.

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa