Answer :
Answer:
2.33 mL
Explanation:
Given data
- Initial pressure (P₁): 1 atm (Standard pressure)
- Initial volume (V₁): 80.0 mL
- Initial temperature (T₁): 273.15 K (Standard temperature)
- Final pressure (P₂): 92.0 atm
- Final volume (V₂): ?
- Final temperature (T₂): 460°C + 273.15 = 733 K
We can find the new volume using the combined gas law.
[tex]\frac{P_1 \times V_1 }{T_1} = \frac{P_2 \times V_2 }{T_2}\\V_2 = \frac{P_1 \times V_1 \times T_2 }{T_1 \times P_2}\\V_2 = \frac{1 atm \times 80.0mL \times 733K }{273.15K \times 92.0atm}=2.33 mL[/tex]
Answer:
The new volume of the gas is 2.33 mL
Explanation:
Step 1: data given
Volume of the fluorine gas = 80.0 mL = 0.080 L
STP = 1 atm and 273 K
The new pressure is 92.0 atm
The new temperature = 460 °C = 733 K
Step 2: Calculate the new volume
P1*V1 / T1 = P2 *V2 / T2
⇒with P1 = the initial pressure of the fluorine gas = 1 atm
⇒with V1 = the initial volume of the fluorine gas = 0.080 L
⇒with T1 = the initial temperature of the fluorine gas = 273 K
⇒with P2 = the new pressure of the gas = 92.0 atm
⇒with V2 = the new volume of the gas = TO BE DETERMINED
⇒with T2 = the new temperature of the gass = 733 K
(1 atm * 0.080 L)/273 K = (92.0 atm * V2) / 733 K
2.93 * 10^-4 = (92.0 atm * V2) / 733 K
0.214769 = (92.0 atm * V2)
V2 = 0.214769 / 92.0
V2 = 0.00233 L = 2.33 mL
The new volume of the gas is 2.33 mL