   Chapter 17, Problem 31E

Chapter
Section
Textbook Problem

# Choose the substance with the larger positional probability in each case.a. 1 mole of H2 (at STP) or 1 mole of H2 (at l00°C, 0.5 atm)b. 1 mole of N2 (at STP) or 1 mole of N2 (at l00 K, 2.0 atm)c. 1 mole of H2O(s) (at 0°C) or 1 mole of H2O(l) (at 20°C)

(a)

Interpretation Introduction

Interpretation: The substance that has larger positional probability is to be identified in each case.

Concept introduction: The substance that can occupy larger volume has the larger positional probability.

The expression for ideal gas law is,

P1V1T1=P2V2T2

Explanation

Explanation

Given

Pressure of H2 at STP is 1atm .

Temperature of H2 at STP is 273K .

Pressure of H2 at 100°C is 0.5atm .

The conversion of degree Celsius (°C) into Kelvin (K) is done as,

T(K)=T(°C)+273

Hence,

The conversion of 100°C into Kelvin is,

T(K)=T(°C)+273T(K)=(100+273)K=373K

Formula

The expression for ideal gas law is,

P1V1T1=P2V2T2

Where,

• P1 is the pressure at STP.
• V1 is the volume at STP.
• T1 is the temperature at STP.
• P2 is the pressure at 100°C .
• V2 is the volume at 100°C .
• T2 is the temperature (100°C)

(b)

Interpretation Introduction

Interpretation: The substance that has larger positional probability is to be identified in each case.

Concept introduction: The substance that can occupy larger volume has the larger positional probability.

The expression for ideal gas law is,

P1V1T1=P2V2T2

(c)

Interpretation Introduction

Interpretation: The substance that has larger positional probability is to be identified in each case.

Concept introduction: The substance that can occupy larger volume has the larger positional probability.

The expression for ideal gas law is,

P1V1T1=P2V2T2

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