Question

A gas with a volume of 525 mL at a temperature of -25°C is heated to 175°C.

What is the new volume, in milliliters, of the gas if pressure and number of moles

are held constant?

Answer 1

Answer:

Volume V2 = 948.13 ml

Explanation:

Given:

Volume V1 = 525 ml

Temperature T1 = -25°C + 273.15

Volume V2 = ?

Temperature T1 = 175°C + 273.15

Computation:

V1 / T1 = V2 / T2

525 / [-25°C + 273.15] = V2 / [175°C + 273.15]

Volume V2 = 948.13 ml

Answer 2

The new volume, in milliliters, of the gas is 948 L

From the question,

We are to determine the new volume of the gas.

From Charles' law which states that the volume of a fixed mass of gas is directly proportional to the temperature (in Kelvin) provided that the pressure remains constant.

That is,

V ∝ T

Then,

V = kT

Therefore, we can write that

$\frac{V_{1} }{T_{1}} = \frac{V_{2}}{T_{2}}$

Where $V_{1}$ is the initial volume

$T_{1}$ is the initial temperature

$V_{2}$ is the final volume

and $T_{2}$ is the final temperature

From the given information,

$V_{1} = 525 \ mL$

$T_{1} = -25 ^{\circ}C = -25 + 273.15 \ K = 248.15 \ K$

$T_{2} = 175 ^{\circ} C = 175 +273.15 \ K =448.15 \ K$

Putting the values into the formula, we get

$\frac{525}{248.15} = \frac{V_{2} }{448.15}$

∴ $V_{2} = \frac{525 \times 448.15}{248.15}$

$V_{2} = \frac{235278.75}{248.15}$

$V_{2} = 948.13 \ L$

V₂ ≅ 948 L

Hence, the new volume, in milliliters, of the gas is 948 L

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