Chapter U3, Problem 20STP

Interpretation Introduction

Interpretation:

The pressure of $2.5\text{ mol}$ gas needs to be calculated under given conditions.

Concept introduction:

Ideal gas is a hypothetical gas for which the assumptions are that the particles have no interaction and occupy negligible space and obeys gas laws.

The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

$\text{PV = nRT}$ - (1)

Where R is Universal gas constant.

Answer to Problem 20STP

The pressure of the gas is $0.93\text{ atm}$.

Explanation of Solution

Given:

Temperature, $\text{T = 32}{\text{.0 }}^{\text{o}}\text{C}$

Volume, $\text{V = 67}\text{.2 L}$

Converting the temperature from degree Celsius to Kelvin:

$\begin{array}{l}\text{T = 32}\text{.0 + 273}\text{.15 }\\ \text{T = 305}\text{.15 K}\end{array}$

Now, in order to determine the pressure of the gas, the equation (1) is rewritten as:

$\text{P = }\frac{\text{nRT}}{\text{V}}$

Value of R used is $0.08206\text{ Latm/molK}$.

Substituting the values in above equation:

$\begin{array}{l}\text{P = }\frac{2.5\text{ mol}\times 0.08206\text{ Latm/molK}\times \text{305}\text{.15 K}}{\text{67}\text{.2 L}}\\ \text{P = }\frac{62.602\text{ Latm}}{\text{67}\text{.2 L}}\\ \text{P = }0.93\text{ atm}\end{array}$

Since, the pressure of the gas is $0.93\text{ atm}$ so, the correct option is (c).