1.16 10-3 mol argon occupies a 69.0 mL container at 44°C, what is the pressure (in torr)?

Answer 1

PV=nRT

P = ?
V = 69 ml = 0.069L
R = 62.36 L •Torr/ K • mole
T = 44º C + 273.15 = 317.15 K
n = 1.16 * 10^-3 mole

P = (1.16*10-3mole) (62.36 L •Torr/ K • mole) (317.15 K) / (0.069L)

= 332.5 torr

Answer 2

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Answer 3

PV = nRT P = nRT/V

T = 44C = 317K
V = 69.0mL = 0.069L
R = 62.4 L-torr/mol-K

P = (1.16x10^-3)(62.4)(317)/(0.069) = 332 torr