Asample of gas at a pressure of 121.59 kpa, a volume of 31 l, and a temperature of 360 k contains how many moles of gas? a. 0.71 mol b. 0.96 mol c. 1.3 mol d. 1.8 mol

C. 1.3 mol

Explanation:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

Given:

P = 121.59 kPa

V = 31 L

T = 360 K

R = 8.3145 L kPa / mol / K

Find: n

n = PV / (RT)

n = (121.59 kPa × 31 L) / (8.3145 L kPa / mol / K × 360 K)

n = (3769.29 L kPa) / (2993.22 L kPa / mol)

n = 1.26 mol

Round to two significant figures, there are 1.3 moles of gas.

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