Chapter 10, Problem 10.77QA

Interpretation Introduction

To find:

a) The moles of a gas in the sample.

b) the probable identity of the gas.

Answer to Problem 10.77QA

Solution:

a) $1.00 mol$

b) Probable gas present in the sample is Helium.

Explanation of Solution

1) Concept:

We are given the volume of the sample at different temperature at constant pressure $1 atm.$ The relation between pressure, volume, temperature, and number of moles is given by ideal gas equation.

So, from the given temperature and volume, we can calculate the number of moles for each set of data and using the grams of sample, we can calculate the molar mass of the sample which will give us an idea about the probable identity of gas.

2) Formula:

$PV=nRT$

3) Given data:

i. $Amount of gas=4.00 g$

ii. $Pressure=1.00 atm$

iii.

$V\left(L\right)$	$T\left(K\right)$
$7.88$	$96$
$3.94$	$48$
$1.97$	$24$
$0.79$	$9.6$
$0.39$	$4.8$

4) Calculations:

The number of moles for each set of data are calculated:

Set 1:

From the first data set, we have $V_1=7.88 L$ and $T_1=96 K$

$PV=nRT$

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

$n=\frac{\left(1.00 atm\right) \times \left(7.88 L\right)}{\left(0.08206 \frac{L.atm}{mol.K}\right)\times \left(96\right) K}$

$n=1.00 mol$

Set 2:

From the second data set, we have $V_2=3.94 L$ and $T_2=48 K$

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

$n=\frac{\left(1.00 atm\right) \times \left(3.94 L\right)}{\left(0.08206 \frac{L.atm}{mol.K}\right)\times \left(48\right) K}$

$n=1.00 mol$

Set 3:

From the third data set, we have $V_3=1.97 L$ and $T_3=24 K$

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

$n=\frac{\left(1.00 atm\right) \times \left(1.97 L\right)}{\left(0.08206 \frac{L.atm}{mol.K}\right)\times \left(24\right) K}$

$n=1.00 mol$

Set 4:

From the fourth data set, we have $V_4=0.79 L$ and $T_4=9.6 K$

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

$n=\frac{\left(1.00 atm\right) \times \left(0.79 L\right)}{\left(0.08206 \frac{L.atm}{mol.K}\right)\times \left(9.6\right) K}$

$n=1.00 mol$

Set 5:

From the fifth data set, we have $V_5=0.39 L$ and $T_5=4.8 K$

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

$n=\frac{\left(1.00 atm\right) \times \left(0.39 L\right)}{\left(0.08206 \frac{L.atm}{mol.K}\right)\times \left(4.8\right) K}$

$n=1.00 mol$

So, from the above calculations, it is clear that number of moles of the sample is $1 mole$. We have grams of the sample and also the moles of it. So, we can calculate the molar mass of the sample as

$Molar mass=\frac{4.00 g}{1.00 mol}$

$Molar mass=4 \frac{g}{mol}$

As molar mass of the unknown gas sample comes out to be $4 \frac{g}{mol}$, the probable identity of the gas is Helium since it has molar mass $4 \frac{g}{mol}.$

Conclusion:

The moles of gas for each set of data are calculated, and using the given mass, the molar mass of the gas is calculated. From this, the identity of the gas is determined.