Respuesta :
Answer:
New volume = 355 mL
Explanation:
Using Ideal gas equation for same mole of gas as
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]
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:
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]
[tex]\frac{{1.54}\times {250}}{296.15}=\frac{{1}\times {V_2}}{273.15}[/tex]
Solving for V₂ , we get:
V₂ = 355 mL
Answer:
The new volume is 388 mL
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
