Question

Question

question
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What volume (in L) will a 32 g sample of butane gas, C4H10(9), occupy at a temperature of 45.0 °C and a pressure of 728
mm Hg?

Answer 1

Step-by-step explanation:

The volume of the gas occupied can be calculated by using the ideal gas equation:

`PV=nRT`

where,

P=pressure of the gas in atm

V=volume of the gas in L.

n=number of moles of the gas

R=0.0821L.atm.mol-1.K-1

T=absolute temperature

To get the volume of the gas, follow the below steps:

1) Calculate the number of moles of gas:

Number of moles of butane=mass of butane given/its molecular mass

`=32g/58.0g/mol\\=0.55mol`

2) Convert temperature into kelvin scale:

T=(45+273)K=318K

3)Convert pressure into atm:

760 mm Hg =1 atm

then,

728 mm Hg=

728 mm Hg x 1 atm /760 mm Hg

=0.957 atm

Substitute all these values in the ideal gas equation to get the volume:

`V=(nRT)/(P) \\V=0.55mol x 0.0821 L.atm.mol-1.K-1 x 318K / 0.957 atm\\V=15.0L`

Answer:

The volume of butane gas is 15.0 L

Answer 2

The volume of the gas is 0.015 m^3.

Step-by-step explanation:

mass, m = 32 g

Temperature, T = 45 °C = 45 + 273 = 318 K

Pressure, P = 728 mm of hg = 0.728 x 13.6 x 1000 x 9.8 = 97027.84 Pa

Atomic mass = 4 x 12 + 10 x 1 = 58 g

Use the ideal gas equation

Let the volume is V.

P V = n R T

`97027.8 * V = (32)/(58)* 8.31 * 318 \\\\V = 0.015 m^3`