Introduction to Chemistry
Introduction to Chemistry
4th Edition
ISBN: 9781259288722
Author: BAUER
Publisher: MCG
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Chapter 9, Problem 160QP

(a)

Interpretation Introduction

Interpretation:

The correct observation is to identified.

(a)

Expert Solution
Check Mark

Explanation of Solution

The ideal gas equation states that pressure P and volume V of a gas is directly proportional to the temperature of the gas T and the number of moles n present in the gas.

PV=nRT …… (1)

Here, R is the universal gas constant having value 0.08206 L atm mol1K1 .

1.25 mol of N2 gas at 25°C and 755 torr is heated to 50.0°C and the pressure changes to 978 torr .

Convert temperature units from degree Celsius to kelvin units:

TK=T°C+273.15

Convert 25°C from degree Celsius to kelvin units:

T1=25+273.15=298.15 K

Convert 50.0°C from degree Celsius to kelvin units:

T2=50+273.15=323.15 K

Covert pressure units from torr to atm as follows:

1 atm=760 torr1 torr=1 atm760 torr

Convert 755 torr units from torr to atm as follows:

755 torr=1 atm760 torr×755 torr=0.99 atm

Convert 978 torr units from torr to atm as follows:

978 torr=1 atm760 torr×978 torr=1.28 atm

Substitute P1 as 0.99 atm , n as 1.25 mol , R as 0.08206 L atm mol1K1 , and T1 as 298.15 K in equation (1):

0.99 atm×V1=1.25 mol×0.08206 L atm mol1K1×298.15 KV1=1.25 mol×0.08206 L atm mol1K1×298.15 K0.99 atm=30.8 L

Substitute P2 as 1.28 atm , n as 1.25 mol , R as 0.08206 L atm mol1K1 , and T1 as 323.15 K in equation (1):

1.28 atm×V2=1.25 mol×0.08206 L atm mol1K1×323.15 KV2=1.25 mol×0.08206 L atm mol1K1×323.15 K1.28 atm=25.8 L

Therefore, volume decreases as pressure increases. Hence, option A is incorrect.

The correct statement is, the volume will decrease because temperature and pressure of the gas increase.

(b)

Interpretation Introduction

Interpretation:

The correct observation is to identified.

(b)

Expert Solution
Check Mark

Explanation of Solution

Substitute P1 as 0.99 atm , n as 1.25 mol , R as 0.08206 L atm mol1K1 , and T1 as 298.15 K in equation (1):

0.99 atm×V1=1.25 mol×0.08206 L atm mol1K1×298.15 KV1=1.25 mol×0.08206 L atm mol1K1×298.15 K0.99 atm=30.8 L

Therefore, option B is incorrect, since the initial volume of the gas is 30.8 L .

The correct statement is, the initial volume is 30.8 L .

(c)

Interpretation Introduction

Interpretation:

The correct observation is to identified.

(c)

Expert Solution
Check Mark

Explanation of Solution

Substitute P2 as 1.28 atm , n as 1.25 mol , R as 0.08206 L atm mol1K1 , and T1 as 323.15 K in equation (1):

1.28 atm×V2=1.25 mol×0.08206 L atm mol1K1×323.15 KV2=1.25 mol×0.08206 L atm mol1K1×323.15 K1.28 atm=25.8 L

Therefore, volume decreases as temperature increase. So, option C is incorrect.

The correct statement is the volume is expected to decrease because the pressure increases by a larger factor than temperature.

(d)

Interpretation Introduction

Interpretation:

The correct observation is to identified.

(d)

Expert Solution
Check Mark

Explanation of Solution

Substitute P2 as 1.28 atm , n as 1.25 mol , R as 0.08206 L atm mol1K1 , and T1 as 323.15 K in equation (1):

1.28 atm×V2=1.25 mol×0.08206 L atm mol1K1×323.15 KV2=1.25 mol×0.08206 L atm mol1K1×323.15 K1.28 atm=25.8 L

So, the final volume is 25.8 L . Option D is correct.

(e)

Interpretation Introduction

Interpretation:

The correct observation is to identified.

(e)

Expert Solution
Check Mark

Explanation of Solution

The initial volume is 30.8 L and the final volume is 25.8 L . Therefore, option E is incorrect.

The correct statement is, the volume of 1 mol of a gas at STP occupies 22.414 L .

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Chapter 9 Solutions

Introduction to Chemistry

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