Chapter 7, Problem 7.30EP

Interpretation Introduction

Interpretation:

Based on Charles’s law, the blanks in the table by using the given condition has to be filled.

Concept Introduction:

Charles’s Law tells that the relationship between temperature and volume of a gas.  The law states that at constant pressure the volume of a fixed amount of gas is directly proportional to the temperature in Kelvin units.  The mathematical expression for the relationship between volume and Kelvin temperature according to the Charles’s law can be shown as follows,

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

  Where,

$\begin{array}{l}{V}_1=Thevolumeofagasatagivenpressure.\\ T_1=TheKelvintemperatureofthegas.\\ V_2=Thevolumeofthegasunderanewsetofconditions.\\ T_2=TheKelvintemperatureofthegasunderanewsetofconditions.\end{array}$

Answer to Problem 7.30EP

Based on Charles’s law, by using the given conditions, the blanks in the table is filled as follows.

	V1	T1	V2	T2
	2.00L	27ºC	4.00L	327ºC
a	3.30L	165ºC	4.05L	265ºC
b	7.90L	270ºC	4.45L	33ºC
c	2.78L	227ºC	4.50L	535ºC
d	4.21L	35ºC	4.351L	45ºC

Explanation of Solution

(a)

Record the given data,

$\begin{array}{l}{V}_1=3.30L\\ T_1=165°Cconvertedto438K\\ V_2=?\\ T_2=265°Cconvertedto538K\end{array}$

Now, substitute the given data in the Charles’s law and do some mathematical calculation to get the answer as follows,

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

$V_2=\frac{3.30L\times 538K}{438K}=4.05L.$

Therefore, the new volume under the new conditions is $4.05L.$

(b)

Record the given data,

$\begin{array}{l}{V}_1=7.90L\\ T_1=270°Cconvertedto543K\\ V_2=4.45L\\ T_2=?\end{array}$

Now, substitute the given data in the Charles’s law and do some mathematical calculation to get the answer as follows,

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

$T_2=\frac{4.45\cancel{L}\times 543K}{7.90\cancel{L}}=306K.$

The obtained temperature is in Kelvin units so convert it into degree Celsius units as follows,

$T-273°C$

$306K-273=33°C$

Therefore, the new temperature in degree Celsius under the new conditions is $33°C.$

(c)

Record the given data,

$\begin{array}{l}{V}_1=?\\ T_1=227°Cconvertedto500K\\ V_2=4.50L\\ T_2=535°Cconvertedto808K\end{array}$

Now, substitute the given data in the Charles’s law and do some mathematical calculation to get the answer as follows,

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

$V_1=\frac{4.50L\times 500K}{808K}=2.78L.$

Therefore, the initial volume under the new conditions is $2.78L.$

 (d)

Record the given data,

$\begin{array}{l}{V}_1=4.21L\\ T_1=?\\ V_2=4.35L\\ T_2=45°Cconvertedto318K\end{array}$

Now, substitute the given data in the Charles’s law and do some mathematical calculation to get the answer as follows,

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

$T_1=\frac{4.21\cancel{L}\times 318K}{4.35\cancel{L}}=308K.$

The obtained temperature is in Kelvin units so convert it into degree Celsius units as follows,

$T-273°C$

$308K-273=35°C$

Therefore, the initial temperature in degree Celsius under the new conditions is $127°C.$