Chapter 7, Problem 7.45EP

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 7.45EP

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

	P	V	n	T
	24.6 atm	1.00 L	1.00mole	27oC
a	32.8 atm	2.00 L	2.00 mole	127oC
b	6.00 atm	10.9 L	4.00 mole	-73oC
c	3.75 atm	3.75 L	0.459 mole	100oC
d	1.22 atm	2.00 L	0.500 mole	-214oC

Explanation of Solution

(a)

Record the given data,

$\begin{array}{l}\text{P = ?}\\ \text{V =2}\text{.00 L}\\ \text{n = 2}\text{.00 mole}\\ \text{T = 127}°\text{C converted to 400K}\end{array}$

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

$PV=nRT$

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

$P=\frac{\left(2.00\cancel{mole}\right)\left(0.0821atm.\cancel{L}/\cancel{mole.K}\right)\left(400\cancel{K}\right)}{\left(2.00\cancel{L}\right)}=32.8atm$

Therefore, the new pressure is $\text{}32.8atm.$

(b)

Record the given data,

$\begin{array}{l}\text{P = 6}\text{.00 atm}\\ \text{V =?}\\ \text{n = 4}\text{.00 mole}\\ \text{T = -73}°\text{C converted to 200K}\end{array}$

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

$PV=nRT$

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

$V=\frac{\left(4.00\cancel{mole}\right)\left(0.0821\cancel{atm}.L/\cancel{mole.K}\right)\left(200\cancel{K}\right)}{\left(6.00\cancel{atm}\right)}=10.9L$

Therefore, the new volume is $10.9L.$

(c)

Record the given data,

$\begin{array}{l}\text{P = 3}\text{.75atm}\\ \text{V =3}\text{.75 L}\\ \text{n = ?}\\ \text{T = 100}°\text{C converted to 373K}\end{array}$

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

$PV=nRT$

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

$n=\frac{\left(3.75\cancel{atm}\right)\left(3.75\cancel{L}\right)}{\left(0.0821\cancel{atm}.\cancel{L}/mole.\cancel{K}\right)\left(373\cancel{K}\right)}=0.459mole$

Therefore, the number of moles is $0.459mole.$

(d)

Record the given data,

$\begin{array}{l}\text{P = 1}\text{.22atm}\\ \text{V =2}\text{.00 L}\\ \text{n = 0}\text{.500 mole}\\ \text{T = }?\end{array}$

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

$PV=nRT$

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

$T=\frac{\left(1.22\cancel{atm}\right)\left(2.00\cancel{L}\right)}{\left(0.0821\cancel{atm}.\cancel{L}/\cancel{mole}.K\right)\left(0.500\cancel{mole}\right)}=59.43K$

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

$T-273$

$59-273=-214°C$

Therefore, the new temperature is $-214°C.$