   Chapter 13, Problem 62QAP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
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# Suppose that 1 . 29 g of argon gas is confined to a volume of 2 . 41 L at 29  ° C . What would be the pressure in the container? What would the pressure become if the temperature were raised to 42  ° C without a change in volume?

Interpretation Introduction

Interpretation:

Pressure in the container containing 1.29 g of argon gas should be calculated.

Concept Introduction:

According to ideal gas equation:

PV=nRT

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature.

The value of Universal gas constant can be taken as 0.082 L atm K1 mol1.

Explanation

Calculation:

The mass of argon gas is 1.29 g, number of moles can be calculated as follows:

n=mM

Molar mass of argon gas is 39.946 g/mol thus,

n=1.29 g39.946 g/mol=0.0323 mol

Now, the pressure of gas can be calculated using the following formula:

P=nRTV

Convert the temperature from K to C as follows:

0 C=273.15 K

Thus,

29 C=(29+273.15) K=302.15 K

Putting the values,

P=(0.0323 mol)(0.082 L atm K1 mol1)(302.15 K)(2.41 L)=0

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