Question

Find the volume of 32g of methane ideal gas at 27degree calcium and 380mm hg

Answer 1

Answer:

hope it works

Explanation:

At first we should convert the pressure from mm Hg into pascal .

380 mm Hg = 0.5 atm

0.5 atm = 0.5 × 10^5

so , P = 5 × 10^4

according ideal gas equation :-

PV = nRT

V = nRT/P

V = (2 × 8.3 × 300) / 5 × 10^4

V ~ 0.1 m3 = 100 litres

Answer 2

We know that, 1 mole of any gas at STP conditions has a volume of 22.4 litres.

STP conditions are: 273 K temperature and 760 mm of Hg

molar mass of methane is 12 + 4 => 16 grams per mole

So, 16 X 2 grams of CH4 is 1 X 2 => 2 moles.

At STP, 2 moles of methane gas will have a volume of 22.4 X 2 => 44.8 L

So, volume of 2 moles of methane at 27 C and 380 mm of Hg = ?

V1 = 44.8 L                                                                    V2 = ?

T1 = 273 K                                                                      T2 = 27 + 273=>300 K

P1 = 760 mm of Hg                                                        P2= 380 mm of Hg

By gas equation:

P1V1/T1 = P2V2/T2

760 X 44.8/273= 380 X V2/300

2 X 44.8/273 = V2/300

2 X 44.8/91 = V2/100

V2= 200 X 44.8/91

V2 = 98.461 litres

Hence, the volume will be 98.46 litres approximately.