Respuesta :
Answer:
Explanation:
Given:
V1 = 200 ml
T1 = 20 °C
= 20 + 273
= 293 K
P1 = 3 atm
P2 = 2 atm
V2 = 400 ml
Using ideal gas equation,
P1 × V1/T1 = P2 × V2/T2
T2 = (2 × 400 × 293)/200 × 3
= 234400/600
= 390.67 K
= 390.67 - 273
= 117.67 °C
The new temperature if the volume increases is 117.67 °C.
Given:
V₁ = 200 ml
T₁ = 20 °C= 20 + 273= 293 K
P₁ = 3 atm
P₂ = 2 atm
V₂ = 400 ml
To find:
T₂ =?
Ideal gas law equation:
PV=nRT
P₁ × V₁/T₁ = P₂ × V₂/T₂
T₂ = (2 × 400 × 293)/200 × 3
T₂= 234400/600
T₂= 390.67 K
T₂= 390.67 - 273
T₂ = 117.67 °C
Thus, the new temperature if the volume increases is 117.67 °C.
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