Respuesta :
Charles law states that volume of a gas is directly proportional to temperature at constant pressure for a fixed amount of gas. the pressure at given conditions is 1.00 atm and the standard pressure at STP conditions too is 1.00 atm therefore pressure remains constant.
V1/T1 = V2/T2
the parameters for the first instance are on the left side and parameters at STP are on the right side of the equation.
T1 - temperature in kelvin - 273 + 30.0 °C = 303 K
T2 - standard temperature - 273 K
V1 - volume - 50.0 L
substituting the values in the equation
50.0 L / 303 K = V / 273 K V = 45.0 L
answer is 2) 45.0 L
V1/T1 = V2/T2
the parameters for the first instance are on the left side and parameters at STP are on the right side of the equation.
T1 - temperature in kelvin - 273 + 30.0 °C = 303 K
T2 - standard temperature - 273 K
V1 - volume - 50.0 L
substituting the values in the equation
50.0 L / 303 K = V / 273 K V = 45.0 L
answer is 2) 45.0 L
Answer: The volume of the gas at STP is 45.0 L. hence the correct answer is option(2).
Explanation:
Initial volume of the gas,[tex]V_1[/tex] = 50.0 L
Initial temperature of the gas[tex]T_1[/tex] =[tex]30.0^oC=30+273 K=303 K[/tex]
[tex]0^oC=273 K[/tex]
Final volume of the gas,[tex]V_2[/tex] = ?
At STP, the value of temperature is 273 K.
Final temperature of the gas[tex]T_2[/tex] = 273 H
Charles' Law: The volume is directly proportional to the temperature of the gas at constant pressure and number of moles.
[tex]V\propto T[/tex] (At constant pressure and number of moles)
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
[tex]V_2=\frac{V_1\times T_2}{T_1}=\frac{50.0 L\times 273 K}{303 K}=45.0495 L\approx 45.0 L[/tex]
The volume of the gas at STP is 45.0 L. hence the correct answer is option(2).