Question

a movable chamber has a volume of 18.5 L (at temperature of 18.5 C) assuming no gas escapes and the pressure remains constant what is the temperature when the chamber has a volume of 19.8 L

Answer 1

Answer:

$39 ^\circ C$

Explanation:

Given,

V₁ = 18.5 L

T₁ = 18.5° C = 273 + 18.5 = 291.5 K

V₂ = 19.8 L

T₂ = ?

Pressure is constant

Using ideal gas equation

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

$\dfrac{18.5}{291.5}=\dfrac{19.8}{T_2}$

$T_2 = 312 K$

$T_2 = 312 -273 =39 ^\circ C$

Answer 2

The temperature when the chamber has a volume of 19.8 L should be considered as the 39 degrees.

Calculation of the temperature:

Since

V₁ = 18.5 L

T₁ = 18.5° C = 273 + 18.5 = 291.5 K

V₂ = 19.8 L

So here the final temperature should be

V1/T1 = V2/T2

18.5 L/291.5 = 19.8 / T2

So T2 = 312 k

= 312 - 273

= 39 degrees

hence, The temperature when the chamber has a volume of 19.8 L should be considered as the 39 degrees.

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