A cylinder of oxygen gas contains 91.3 g O_2 if the volume of the cylinder is 8.

Question

A cylinder of oxygen gas contains 91.3 g O_2 if the volume of the cylinder is 8.58 L, what is the pressure of the O_2 if the gas temperature is 21 degree C? In an experiment, you fill a heavy-walled 6.00-LAask with methane gas, CH_4. If the flask contains 7.13 g of methane at 19 degree C, what is the gas pressure? An experiment calls for 3.70 mol of chlorine, CI_2. What volume will this be if the gas volume is measured at 36 degree C and 3.30 atm? According to your calculations, a reaction should yield 5.67 g of oxygen, O_2. What do you expect the volume to be at 23 degree C and 0.894 atm?

Explanation / Answer

PV = nRT

where:

P is the pressure of the gas

V is the volume of the gas

n is the amount of substance of gas (in moles)

R is gas constant = = 0.0821 L atm K-1 Mol-1

T is the absolute temperature of the gas.

Question 5.57

P = ?

V= 8.58 L

n = 91.3 /32 = 2.853 Mole

T = 273+21 = 294 K

P = 2.853 x 0.0821 x 294 / 8.58 = 8.02 atm

P = 8.02 atm

Question 5.57

P = ?

V= 6.00 L

n = 7.13 /16 = 0.445 Mole

T = 273+ 19 = 292 K

P = 0.445 x 0.0821 x 292 / 6 = 1.78 atm

P = 1.78 atm

Question 5.59

P = 3.3 atm

V = ?

n = 3.7 mole

T = 273 + 36 = 309 K

V = 3.7 x 0.0821 x 309 / 3.3 = 28.44 Liter

V = 28.44 Liter

Question 5.60

P = 0.894 atm

V= ?

n = 5.67/32 = 0.1771 Mole

T = 273 +23 = 296 K

V = 0.1771 x 0.0821 x 296 / 0.894 = 4.816 Liter

V = 4.816 Liter

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