Chapter 13, Problem 35P

To determine

To Find: The pressure of the argon gas at a given temperature using ideal gas law.

Answer to Problem 35P

The pressure of the argon gas at given temperature using Ideal gas law $2.40\times {10}^8\text{Pa}$ .

Explanation of Solution

Given:

Temperature= $\mathrm{T}\text{=}\text{385}\text{K}$

Mass of the argon gas= $\mathrm{m}\text{=}\text{1}0\text{5}\text{kg}$

The volume of the container is $\mathrm{V}\text{=}\text{35}\text{L}$

Formula used:

Write the expression for the ideal gas equation.

  $\begin{array}{l}\mathrm{P}\mathrm{V}=\mathrm{n}\mathrm{R}\mathrm{T}\\ \mathrm{P}=\frac{\mathrm{n}\mathrm{R}\mathrm{T}}{\mathrm{V}}\end{array}$

Where,

R is the gas constant

P = Pressure of the gas

V =Volume of the gas

n = no of moles

T = Temperature of the gas

Calculation:

Consider that the argon gas is the ideal gas. To calculate the no of moles of argon we use the atomic weight of the gas. After no of moles use the ideal gas law to find the pressure of the gas.

  $\begin{array}{c}\mathrm{n}=(105\text{kg})\frac{1\text{mol}\text{Ar}}{39.95\times {19}^{-3}\text{kg}}\\ =2628\text{mol}\\ \mathrm{P}\mathrm{V}=\mathrm{n}\mathrm{R}\mathrm{T}\\ \mathrm{P}=\frac{\mathrm{n}\mathrm{R}\mathrm{T}}{\mathrm{V}}=\frac{(2628\text{mol})(8.315\text{J/mol}\text{.K})(385\text{K})}{(35.0\text{L})(1\times {10}^{-3}{\text{m}}^{\text{3}}\text{/L})}\\ =2.40\times {10}^8\text{Pa}\end{array}$

Conclusion:

The pressure of the argon gas is $2.40\times {10}^8\text{Pa}$ .