What is the temperature of a gas that is expanded from 3.75 l at 37 degrees celsius to 5.6 l?

190 °C  

Step-by-step explanation:

The pressure is constant, so this looks like a case where we can use Charles’ Law:  

V₁/T₁ = V₂/T₂      Invert both sides of the equation.  

T₁/V₁ = T₂/V₂      Multiply each side by V₂

T₂ = T₁ × V₂/V₁

V₁ = 3.75 L; T₁ = (37 + 273.15) K = 310.15 K  

V₂ = 5.6 L;   T₂ = ?  

T₂ = 310.15 × 5.6/3.75

T₂ = 310.15 × 1.49

T₂ = 463 K

t₂ = 463 – 273.15

t₂ = 190 °C

answer is A

hope this helps