INTOR TO CHEMISTRY LLF
INTOR TO CHEMISTRY LLF
5th Edition
ISBN: 9781264501731
Author: BAUER
Publisher: MCG
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Chapter 9, Problem 132QP
Interpretation Introduction

Interpretation:

The total pressure of mixture of gas and partial pressure of argon gas and nitrogen gas in the tank at 45°C are to be determined.

Concept Introduction:

The number of moles n present in the gas container is equal to the ratio of amount of gas m present in the container to the molar mass of the gas M .

n=mM … (1)

Ideal gas law is followed by ideal gases. It states that pressure P and volume V of the gas are directly proportional to the number of moles n present in the gas and the temperature T of the gas.

PVnTPV=nRT … (2)

Where, R is universal gas constant.

Dalton’s law of partial pressure states that the total pressure of the mixture of gases is equal to the sum of partial pressure of gases present in the mixture of gases.

PTotal=P1+P2+P3+..........+Pn

Expert Solution & Answer
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Answer to Problem 132QP

Solution:

The partial pressure of argon gas and nitrogen gas in the tank are 21.5 atm and 33.6 atm , respectively. The total pressure of the mixture of gases in the tank is 55.1 atm .

Explanation of Solution

Given Information: The volume of the tank is 2.00 L . Tank contains 72.0 g N2 and 66.0g Ar .

The tank contains a mixture of gases 72.0 g N2 and 66.0g Ar . The moles of each gas present in the tank can be calculated by using equation (1). The molar mass of Ar gas is 39.95 g mol1 . Substitute mAr as 66.0g and MAr as 39.95 g mol1 in equation (1) to determine the no. of moles of argon gas.

nAr=mArMAr=66.0 g39.95 g mol1=1.65 mol

The molar mass of N2 gas is 28 g mol1 . Substitute mN2 as 72.0 g and MN2 as 28 g mol1 in equation (1) to determine the number of moles of nitrogen gas.

nN2=mN2MN2=72.0 g28 g mol1=2.57mol

The moles of argon gas are 1.65 mol and the moles of nitrogen gas are 2.57mol in the tank. The temperature of the mixture of gas is 45.0°C and the volume of the gas is 2 L . Pressure exerted by moles of argon gas is determined by equation (2).

Convert temperature units from Celsius to Kelvin.

TK=T°C+273.15

For 45.0°C ,

T=45.0°C+273.15=318.2 K

Substitute T as 318.2 K , n as 1.65 mol , V as 2L , and R as 0.08206 L atm mol1K1 in equation (2) to determine the partial pressure of argon gas.

PAr×2 L=1.65 mol×0.08206 L atm mol1K1×318.2 KPAr=1.65 mol×0.08206 L atm mol1K1×318.2 K2 L=21.5 atm

Therefore, the partial pressure of argon gas in the tank is 21.5 atm . Now, to determine the partial pressure of nitrogen gas, substitute T as 318.2 K , n as 2.57mol , V as 2L , and R as 0.08206 L atm mol1K1 in equation (2).

PN2×2 L=2.57 mol×0.08206 L atm mol1K1×318.2 KPN2=2.57 mol×0.08206 L atm mol1K1×318.2 K2 L=33.6 atm

Therefore, the partial pressure of nitrogen gas in the tank is 33.6 atm .

The total pressure of the mixture of gases in the tank is the sum of the partial pressure of argon gas and nitrogen gas according to Dalton’s law. Therefore, the total pressure of gases is,

PT=PAr+PN2=21.5 atm+33.6 atm=55.1 atm

Therefore, the total pressure of the mixture of gases in the tank is 55.1 atm .

Conclusion

The partial pressure of argon gas and nitrogen gas in the tank are 21.5 atm and 33.6 atm . The total pressure of the mixture of gases in the tank is 55.1 atm .

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Chapter 9 Solutions

INTOR TO CHEMISTRY LLF

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