Chapter 13, Problem 39P

To determine

To Find: The new pressure

Answer to Problem 39P

3.427 atm.

Explanation of Solution

Given:

  $\begin{array}{l}{P}_1=2.45\text{atm}\\ V_1=61.5\text{L}\\ V_2=48.8\text{L}\\ T_1={18}^0\text{C}=18+273\text{ K}=291\text{ K}\\ T_2={50}^0\text{C}=50+273\text{ K}=323\text{ K}\\ P_2=?\end{array}$

Formula used:

From ideal gas equation:

  $\begin{array}{l}PV=nRT\\ \frac{PV}{T}=nR\end{array}$

Where,

P = Pressure of the gas

V = Volume of the gas

T = temperature of the gas

n = no of moles

Calculation:

Suppose oxygen is the ideal gas. Means gas is constant and the value is constant of $\frac{PV}{T}$

So,

  $\begin{array}{l}\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}\\ P_2=P_1\frac{V_1{T}_2}{V_2{T}_1}\end{array}$

  $\begin{array}{c}{P}_2=(\text{2}\text{.45}\text{atm)}\frac{\left(\text{61}\text{.5}\text{L}\right)\left(323\text{ K}\right)}{\left(\text{48}\text{.8}\text{L}\right)\left(291\text{ K}\right)}\\ =3.427\text{atm}\end{array}$

Conclusion:

The change in the pressure of the oxygen gas using the ideal gas equation is 3.427 atm.