Question

What is the volume of 3.5 kg of Argon at 25°C and 550 kPa? Express in m3 and ft3

Answer 1

Answer: The volume of argon will be
`0.394m^3` and
`14ft^3`

Step-by-step explanation:

According to the ideal gas equation:

`PV=nRT`

P = Pressure of the gas = 550 kPa = 5.43 atm (1 kPa= 0.0098 atm)

V= Volume of the gas = ?

T= Temperature of the gas = 25°C = 298 K
`0^00C=273K`

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas

Moles of gas=
`\frac{\text{ given mass}}{\text{ molar mass}}= (3.5* 1000g)/(40g/mole)=87.5moles`

`V=(nRT)/(P)=(87.5* 0.0821* 298)/(5.43)=394L`

Thus volume of argon will be 394 Liters

1. Converting L to
`m^3`

`1L=0.001m^3`

`394L=(0.001)/(1)* 394=0.394m^3`

Thus volume of argon will be
`0.394m^3`

2. Converting L to
`ft^3`

`1L=0.035ft^3`

`394L=(0.035)/(1)* 394=14ft^3`

Thus volume of argon will be
`14ft^3`

Answer 2

Answer: The volume of given amount of argon is
`0.394m^3` and
`13.91ft^3`

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Given mass of argon = 3.5 kg = 3500 g (Conversion factor: 1 kg = 1000 g)

Molar mass of argon = 40 g/mol

Putting values in above equation, we get:

`\text{Moles of argon}=(3500g)/(40g/mol)=87.5mol`

To calculate the volume of gas, we use the equation given by ideal gas equation:

`PV=nRT`

where,

P = pressure of the gas = 550 kPa

V = Volume of gas = ?

n = number of moles of argon = 87.5 moles

R = Gas constant =
`8.31\text{L kPa }mol^(-1)K^(-1)`

T = temperature of the gas =
`25^oC=[25+273]=298K`

Putting values in above equation, we get:

`550kPa* V=87.5mol* 8.31\text{L kPa }mol^(-1)K^(-1)* 298K\\\\V=394L`

Converting volume from liters to cubic meters and cubic foot, we use the conversion factor:

`1m^3=1000L\\\\1ft^3=28.32L`

Converting the given volume into cubic meters:

`\Rightarrow 394L* ((1m^3)/(1000L))=0.394m^3`

Converting the given volume into cubic foot:

`\Rightarrow 394L* ((1ft^3)/(28.32L))=13.91ft^3`

Hence, the volume of given amount of argon is
`0.394m^3` and
`13.91ft^3`