21.4 L of gas at 42.3 °C and 681 mm Hg are heated to 221 °C, and the pressure is changed to 248 mm Hg. What is the new volume?

V₂ = 90.42 L

Explanation:

Given data:

Initial volume = 21.4 L

Initial pressure = 681 mmHg (681/760=0.89 atm)

Initial temperature = 42.3 °C (42.3 +273 = 315.3 K)

Final temperature = 221°C (221+273 = 494 K)

Final volume = ?

Final pressure = 248 mmHg(248/760 =0.33 atm)

Formula:

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂  

V₂ = P₁V₁T₂/T₁P₂

V₂ = 0.89 atm × 21.4 L × 494 K / 315.3 K × 0.33 atm

V₂ = 9408.72 L /104.05

V₂ = 90.42 L

the answer is either b or c