Question

An aerosol can filled with propane gas occupies a volume of 250 mL at 23 C and 1.54 atm. if the conditions are changed to STP, what is the new volume in the can?

Answer 1

The new volume is 388 mL

Step-by-step explanation:

Step 1: Data given

Volume = 250 mL = 0.250 L

Temperature = 23 °C = 296 Kelvin

Pressure = 1.54 atm

Step 2: Calculate number of moles

p*V =n*R*T

n = (p*V)/(R*T)

⇒ n = the number of moles = ?

⇒ V = the volume: the gas occupies = 0.250 L

⇒ The pressure of propane gas = 1.54 atm

⇒ Temperature = 23.0 °C = 296 Kelvin

⇒ R = The gas constant = 0.08206 L*atm/K*mol

n = (1.54 * 0.205 ) / ( 0.08206 * 296 )

n = 0.013 moles

Step 3: Calculate new volume

(P1*V1)/T1 = (P2*V2)/T2

⇒ with P1 = the initial pressure = 1.54 atm

⇒ with V1 = the initial volume = 0.250 L

⇒ with T1 = the initial temperature = 296 K

⇒ with P2 = 1.00 atm

⇒ with V2 = The new volume at STP

⇒ with T2 = 298 Kelvin

(1.54*0.250)/296 = (1.00 * V2)/298

V = 0.388 L = 388 mL

The new volume is 388 mL

Answer 2

New volume = 355 mL

Step-by-step explanation:

Using Ideal gas equation for same mole of gas as

`\frac {{P_1}* {V_1}}{T_1}=\frac {{P_2}* {V_2}}{T_2}`

Given ,

V₁ = 250 mL

V₂ = ?

P₁ = 1.54 atm

T₁ = 23 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (23 + 273.15) K = 296.15 K

At STP,

P₂ = 1 atm

T₂ = 273.15 K

Using above equation as:

`\frac {{P_1}* {V_1}}{T_1}=\frac {{P_2}* {V_2}}{T_2}`

`\frac{{1.54}* {250}}{296.15}=\frac{{1}* {V_2}}{273.15}`

Solving for V₂ , we get:

V₂ = 355 mL