Question

A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mm Hg. The flask is opened and more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C. There are now ________ mol of gas in the flask.

Answer 1

Answer: There are now 2.07 moles of gas in the flask.

Step-by-step explanation:

`PV=nRT`

P= Pressure of the gas = 697 mmHg = 0.92 atm (760 mmHg= 1 atm)

V= Volume of gas = volume of container = ?

n = number of moles = 1.9

T = Temperature of the gas = 21°C=(21+273)K= 294 K (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

`V=(nRT)/(P)=(1.9* 0.0821 * 294)/(0.92)=49.8L`

When more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C, but the volume remains same.Thus again using ideal gas equation to find number of moles.

`PV=nRT`

P= Pressure of the gas = 775 mmHg = 1.02 atm (760 mmHg= 1 atm)

V= Volume of gas = volume of container = 49.8 L

n = number of moles = ?

T = Temperature of the gas = 26°C=(26+273)K= 299 K (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

`n=(PV)/(RT)=(1.02* 49.8)/(0.0821* 299)=2.07moles`

Thus the now the container contains 2.07 moles.