Chapter 13, Problem 56A

Interpretation Introduction

Interpretation:

The volume of the gas in the weather balloon is to be calculated when pressure and temperature of the gas are given at two conditions.

Concept introduction:

Ideal gas equation helps to calculate volume of a gas at given temperature and pressure. According to this law,

$\mathrm{PV}=nRT$

Where

$\mathrm{P}$ is pressure of the gas

$\mathrm{V}$ is volume of the gas

$\mathrm{n}$ is the number of moles of the gas

$\mathrm{R}$ is the ideal gas constant

$\mathrm{T}$ is the temperature of the gas

Answer to Problem 56A

The volume of the gas is $5.91\times {10}^{4}L$

Explanation of Solution

Ideal gas equation helps to calculate volume of a gas at given temperature and pressure. According to this law,

$\mathrm{PV}=nRT$

Or

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Given data is:

$\begin{array}{l}{P}_1=0.995atm\\ P_2=0.720atm\\ V_1=5\times {10}^{6}mL\\ T_1=32°C\\ T_1(K)=T_1(°C)+273\\ =32+273\\ =305K\\ T_2=-12°C\\ =-12+273\\ =261K\end{array}$

Now substituting the values into formula,

$\begin{array}{c}{V}_2=\frac{P_1{V}_1{T}_2}{P_2{T}_1}\\ =\frac{0.995\times 5\times {10}^4\times 261}{0.720\times 305}\\ =5.91\times {10}^{4}L\end{array}$

Therefore volume of the gas is $5.91\times {10}^{4}L$

Conclusion

The volume of the gas is $5.91\times {10}^{4}L$.