At what temperature will 0.82 mol Cl2 exert a pressure of 1.78 atm at a volume of 2.71 L?

Answer 1

Use the ideal gas law:

PV = nRT

P = pressure in atm = 1.78 atm

V = volume in L = 2.71 L

n = moles = 0.82 mol

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin =?

Solving for T, we have...

T = PV/nR = (1.78)(2.71) / 0.82)(0.0821)

T = 71.7K = 72 K (2 sig. figs.)

If you want the temperature in ºC, just add 273 = 345ºC = 350ºC (2 sig. figs.)

Answer 2

PV = nRT

1.78 atm(2.71 L) = 0.82 mol(0.08206 L·atm/K·mol)(T)

Solve for T. (Divide both sides of the equation by n and R).