Question

A sample of argon gas has a volume of 735 mL at a pressure of 1.20 atm and a temperature of 112 degrees Celsius. What is the final volume of the gas, in milliliters when the pressure and temperature of the gas sample are changed to the following, if the amount of gas does not change?

A. 658 mmHg and 281 K

B. 0.55 atm and 75 degrees Celsius

C. 15.4 atm and -15 degrees Celsius

Answer 1

Let us find out the moles of argon present

PV = nRT

P = 1.2 atm V = 0.735 L n = ? R = 0.0821 L atm K-1 Mol-1 T = 273.15 + 112 = 385.15

n = 1.2 x 0.735 / 0.0821 x 385.15 = 0.02789 Mole

A. 658 mmHg and 281 K

658 mm Hg = 658 /760 = 0.8658 atm

P = 0.8658  atm V = ? n = 0.02789 Mole R = 0.0821 L atm K-1 Mol-1 T = 281

V = 0.02789 x 0.0821 x 281 / 0.8658 = 0.743 Liter

0.55 atm and 75 degrees Celsius

P = 0.55 atm V = ? n = 0.02789 Mol R = 0.0821 L atm K-1 Mol-1 T = 273.15 + 75 = 348.15 K

V = 0.02789 x 0.0821 x 348.15 / 0.55 = 1.449 Liter

15.4 atm and -15 degrees Celsius

P = 15.4 atm V = ? n = 0.02789 Mol R = 0.0821 L atm K-1 Mol-1 T = 273.15 - 15 = 258.15 K

V = 0.02789 x 0.0821 x 258.15 / 15.4 = 0.038387 Liter