Chapter U3, Problem 7RE

Interpretation Introduction

Interpretation: The new pressure of the propane gas needs to be determined.

Concept Introduction:

The ideal gas equation is represented as follows:

  $PV=nRT$

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

Answer to Problem 7RE

2.05 atm.

Explanation of Solution

Given Information:

The initial pressure and temperature of the propane gas is 2.00 atm and 30 ^oC respectively. The temperature increased to 37 ^oC.

Calculation:

First convert the given temperature to K as follows:

  $0{}^{\text{o}}\text{C}=\text{273}\text{.15 K}$

Thus,

  $30{}^{\text{o}}\text{C}=30+\text{273}\text{.15 K}=\text{303}\text{.15 K}$

Thus,

  $T_1=303.15\text{ K}$

Also,

  $37{}^{\text{o}}\text{C}=37+\text{273}\text{.15 K}=\text{310}\text{.15 K}$

Thus,

  $T_2=310.15\text{ K}$

The ideal gas equation is as follows:

  $PV=nRT$

The volume of the tank remains the same and the number of moles of propane gas also remains the same thus, for constant volume and number of moles,

  $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

On rearranging,

  $P_2=\frac{P_1}{T_1}\times T_2$

Putting the values,

  $\begin{array}{l}{P}_2=\frac{P_1}{T_1}\times T_2\\ =\frac{2\text{ atm}}{303.15\text{ K}}\times 310.15\text{ K}\\ =\text{2}\text{.05 atm}\end{array}$

Conclusion

Thus, the new pressure of the propane gas is 2.05 atm.