Chapter 13, Problem 42A

Interpretation Introduction

Interpretation: -Volume of a gas filled in the balloon needs to be calculated at STP (standard temperature pressure).

Concept Introduction: - Volume of a gas at different temperatures can be calculated by using combined gas law

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

Answer to Problem 42A

Volume of gas at STP is $45.18L$

Explanation of Solution

If state of a gas is represented by ${\text{P}}_{\text{1}}{\text{, V}}_{\text{1}}{\text{ and T}}_{\text{1}}$ and it changes to ${\text{P}}_2{\text{, V}}_2{\text{ and T}}_{\text{2}}$ , then relationship is given by combined gas law.

According to this law,

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

Any one of the variables can be find out if other fine variable is known.

For calculating volume at STP,

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

  $\frac{P_1{V}_1{T}_2}{T_1{P}_2}=V_2(STP)$

  $\begin{array}{l}{P}_1=742Torr=\frac{742}{760}=0.97atm\\ V_1=50.0L\\ T_1=20^{0}C+273=293K\end{array}$

  $\begin{array}{l}{P}_2=1atm\\ T_2=273K\\ V_2=?\end{array}$ {at standard temperature and pressure}

  $\begin{array}{l}{V}_2=\frac{P_1{V}_1{T}_2}{T_1{P}_2}\\ V_2=\frac{0.97atm\times 50.0L\times 273K}{293K\times 1atm}\\ V_2=\frac{097\times 50.0L\times 273}{293}\\ V_2=\frac{13240.5}{293}=45.18L\end{array}$

Volume of gas at STP is $45.18L$

Conclusion

Volume of gas at STP is $45.18L$