Question

A sample of he gas (2.35 mol) occupies 57.9 l at 300.0 k and 1.00 atm. the volume of this sample is _____ l at 423 k and 1.00 atm.

Answer 1

The  volume  of  the sample  at 423 k  and 1.00 atm  is calculated using  Charles  law  formula   that is V1/T1 =V2/T2 since  the  pressure is  constant

V1=75.9  L
T1=  300 K
V2=?
T2=423 k

by   making the V2 the  subject  of  the formula

V2=T2V1/T1
V2=  57.9 L  x 423 K/300 K=  81.64 L  is the volume of the  sample

Answer 2

The ideal gas relates to the pressure, temperature, pressure, and mole of the gas. The volume of the sample at 423 K and 1 atm is 81.64L.

What is Charles's law?

Charles law gave the relation between the temperature and the volume of gas. According to him, the volume of the gas is directly proportional to the temperature of the gas.

Given,

Initial volume = 75.9 L

Initial temperature = 300 K

Final volume = ?

Final temperature = 423 K

The relation between the temperature and the volume of the two gases is shown as:

$\rm \dfrac{V_{1}}{T_{1}} = \rm \dfrac{V_{2}}{T_{2}}$

Substituting values in the above equation:

$\begin{aligned}\rm V_{2}& = \rm \dfrac{T_{2}V_{1}}{T_{1}}\\\\&= \dfrac{57.9\times 423}{300}\\\\&= 81.64 \;\rm L \end{aligned}$

Therefore, 81.64 L is the final volume at 423 K.

Learn more about Charles law here:

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