Answer:
130.165636364°C
Explanation:
P = Pressure
V = Volume
n = Number of moles
R = Gas constant = 0.082 L atm/mol K
From ideal gas law we have
[tex]PV=nRT\\\Rightarrow n=\dfrac{PV}{RT}\\\Rightarrow n=\dfrac{22\times 25}{0.082\times (25+273.15)}\\\Rightarrow n=22.496451696\ moles[/tex]
[tex]PV=nRT\\\Rightarrow T=\dfrac{PV}{nR}\\\Rightarrow T=\dfrac{48\times 15.5}{22.496451696\times 0.082}\\\Rightarrow T=403.315636364\ K[/tex]
The initial temperature is [tex]403.315636364-273.15=130.165636364\ ^{\circ}C[/tex]
Answer: 130 degrees Celsius
Explanation:
P1V1 / T1 = P2V2 / T2
Let the subscript 1 represent the initial 15.5L of gas and the subscript 2 represent the gas at the final volume of 25L. Rewrite the temperature in Kelvin by adding 273.
P1=48atm, V1=15.5L, P2=22atm, V2=25L, T2=298K, and T1 is unknown.
Substitute the known values into the combined gas law equation.
P1V1 / T1=P2V2 / T2 → (48atm)(15.5L) / T1 = (22atm)(25L) / 298K
Solve the equation for T1, simplify, and round to the nearest degree.
T1 = P1V1T2 / P2V2
T1 = (48atm)(15.5L)(298K)(22atm)(25L)
T1 = 403K
To get the temperature in Celsius, subtract 273.
T1 = 130∘C
