Chapter 7, Problem 7.46EP

Interpretation Introduction

Interpretation:

The blanks in the each row of the table by using ideal gas law at given set of conditions has to be filled.

Concept Introduction:

Ideal gas law provides the relationship between the four gas variables temperature, pressure, volume, and molar amount of a gaseous substance at specified set of conditions.  The mathematical expression for ideal gas law can be shown below,

$\begin{array}{l}PV=nRT\\ \text{Where,}\\ P=\text{Pressure}\text{}\text{of}\text{gas}\\ V=\text{Volume}\text{}\text{of}\text{gas}\\ T=\text{Temperature}\text{of}\text{gas}\\ n=\text{Number}\text{of}\text{moles}\text{of}\text{gas}\\ R=\text{Ideal}\text{gas}\text{constant}\\ \end{array}$

Answer to Problem 7.46EP

At given conditions the blanks in the table is filled by using ideal gas law is,

	P	V	n	T
	10.9 atm	3.00 L	1.00mole	127oC
a	22.4 atm	3.30 L	3.00 mole	27oC
b	7.10 atm	8.00 L	2.00 mole	73oC
c	2.50 atm	2.50 L	0.761mole	-173oC
d	5.00 atm	10.0 L	2.00 mole	32oC

Explanation of Solution

(a)

Record the given data,

$\begin{array}{l}\text{P = 22}\text{.4 atm}\\ \text{V = ?}\\ \text{n = 3}\text{.00 mole}\\ \text{T = 27}°\text{C converted to 300K}\end{array}$

Now, substitute these values in ideal gas law and do some mathematical calculation as shown in below,

$PV=nRT$

$V=\frac{nRT}{P}$

$V=\frac{\left(3.00\cancel{mole}\right)\left(0.0821\cancel{atm}.L/\cancel{mole.K}\right)\left(300\cancel{K}\right)}{\left(22.4\cancel{atm}\right)}=3.30L$

Therefore, the new volume is $3.30L.$

(b)

Record the given data,

$\begin{array}{l}\text{P = ?}\\ \text{V = 8}\text{.00L}\\ \text{n = 2}\text{.00 mole}\\ \text{T = 73}°\text{C converted to 346K}\end{array}$

Now, substitute these values in ideal gas law and do some mathematical calculation as shown in below,

$PV=nRT$

$P=\frac{nRT}{V}$

$P=\frac{\left(2.00\cancel{mole}\right)\left(0.0821atm.\cancel{L}/\cancel{mole.K}\right)\left(346\cancel{K}\right)}{\left(8.00\cancel{L}\right)}=7.10atm$

Therefore, the new pressure is $7.10atm.$

(c)

Record the given data,

$\begin{array}{l}\text{P = 2}\text{.50atm}\\ \text{V =2}\text{.50 L}\\ \text{n = ?}\\ \text{T = -173}°\text{C converted to 100K}\end{array}$

Now, substitute these values in ideal gas law and do some mathematical calculation as shown in below,

$PV=nRT$

$n=\frac{PV}{RT}$

$n=\frac{\left(2.50\cancel{atm}\right)\left(2.50\cancel{L}\right)}{\left(0.0821\cancel{atm}.\cancel{L}/mole.\cancel{K}\right)\left(100\cancel{K}\right)}=\text{0}\text{.761}mole$

Therefore, the number of moles is $\text{0}\text{.761}mole.$

(d)

Record the given data,

$\begin{array}{l}\text{P = 5}\text{.00 atm}\\ \text{V =10}\text{.0 L}\\ \text{n = 2}\text{.00 mole}\\ \text{T = }?\end{array}$

Now, substitute these values in ideal gas law and do some mathematical calculation as shown in below,

$PV=nRT$

$T=\frac{PV}{nR}$

$T=\frac{\left(5.00\cancel{atm}\right)\left(10.0\cancel{L}\right)}{\left(0.0821\cancel{atm}.\cancel{L}/\cancel{mole}.K\right)\left(2.00\cancel{mole}\right)}=305K$

Now, convert the obtained Kelvin temperature into degree Celsius temperature by using the following equation,

$T-273$

$305-273=32°C$

Therefore, the new temperature is $32°C.$