Question

Assigned as ir in an internal combustion engine is at 440'F and 150 psía, with a volume of 10 in3. what is the mass of air b,) ou may assume ideal gas

Answer 1

Answer: The mass of air is 1.18 g

Step-by-step explanation:

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

`PV=nRT`

Or,

`PV=(m)/(M)RT`

where,

P = pressure of the gas = 150 psia = 10.2 atm (Conversion factor: 1 psia = 0.068 atm)

V = Volume of gas =
`10in^3=0.164L` (Conversion factor:
`1in^3=0.0164L` )

m = mass of air = ?

M = Average molar mass of air = 28.97 g/mol

R = Gas constant =
`0.0820\text{ L atm }mol^(-1)K^(-1)`

T = temperature of the gas =
`440^oF=499.817K` (Conversion factor:
`(T(K)-273.15)=(T(^oF)-32)* (5)/(9)` )

Putting values in above equation, we get:

`10.2atm* 0.164L=(m)/(28.97g/mol)* 0.0820\text{ L atm }mol^(-1)K^(-1)* 499.817K\\\\m=1.18g`

Hence, the mass of air is 1.18 g