Question

f a gas has an initial volume of 10L at a temperature of 350K and it is then heated to 680K, what is the new volume?

Answer 1

Answer:

V1 = 10 L

V2 = x L

T1 = 350 K

T2 = 680 K

And you are looking for the second Volume

 the image shows the work and the answer is going to be 19.4 L or just 19 L

Answer 2

Hence, 19.4 L is the new volume of gas when the initial volume of 10L at a temperature of 350K and is then heated to 680K.

What is the ideal gas equation?

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases.

An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given data:

$V_1$ = 10 L

$V_2$ =?

$T_1$ = 350 K

$T_2$= 680 K

Now using equation:

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

$\frac{10 L}{350 K} = \frac{V_2}{680 K}$

$V_2$ = 19.4 L

Hence, 19.4 L is the new volume of gas when the initial volume of 10L at a temperature of 350K and is then heated to 680K.

Learn more about the ideal gas equation here:

https://brainly.com/question/18772359

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