A 350.0 mL sample of gas has a temperature of 22.7 C and a pressure of 0.914 atm. What temperature would be needed for the same amount of gas to fit into a 220.0 mL flask at standard pressure?

T₂  = 203.4 K

Explanation:

Given data:

Initial volume = 350.0 mL

Initial temperature = 22.7 °C

Initial pressure = 0.914 atm

Final temperature = ?

Final volume = 220.0 mL

Final pressure = standard =  1 atm

Solution:

First of all we will convert the temperature into kelvin.

22.7 °C + 273 = 295.7 K

Formula:  

P₁V₁/T₁ = P₂V₂/T₂

T₂  =  P₂V₂T₁  /P₁V₁

T₂  = 1 atm × 220.0 mL × 295.7 K /  0.914 atm  × 350.0 mL

T₂ = 65054 atm.mL.k /319.9 atm.mL

T₂  = 203.4 K

answer; protein;

they are both synthetic elements

explanation:

d