Chapter 1, Problem 1A.1ST

Interpretation Introduction

Interpretation: The temperature that would result in the same given sample exerting a pressure of $300\text{atm}$ has to be stated.

Concept introduction: The perfect gas equation of state is consistent with Boyle’s law, Charles’s law and also with Avogadro’s law.  The perfect gas law is also known as ideal gas equation.  The expression that is used to show the perfect gas equation of state is given as,

    $pV=nRT$

Answer to Problem 1A.1ST

The temperature that would result in the same given sample exerting a pressure of $300\text{atm}$ is $\underset{\_}{900K}$.

Explanation of Solution

It is given that the sample of nitrogen gas has constant volume.

The given initial pressure of the sample $\left(p_1\right)$ is $100\text{atm}$.

The given final pressure of the sample $\left(p_2\right)$ is $300\text{atm}$.

The given initial temperature of the sample $\left(T_1\right)$ is $300\text{K}$.

The perfect gas equation that is used for the two sets of values for the given sample of nitrogen is given as,

    $\frac{p_1{V}_1}{n_1{T}_1}=\frac{p_2{V}_2}{n_2{T}_2}$        (1)

Where,

• $p_1$ is the initial pressure of the gas.
• $p_2$ is the final pressure of the gas.
• $V_1$ is the initial volume of the gas.
• $V_2$ is the final volume of the gas.
• $T_1$ is the initial temperature of the gas.
• $T_2$ is the final temperature of the gas.
• $n_1$ is the initial number of moles of the gas.
• $n_2$ is the final number of moles of the gas.

As the given sample of nitrogen gas behaves as the perfect gas, thus, the initial and final number of moles of the gas is same.

Therefore, for given sample of nitrogen gas $V_1=V_2$ and $n_1=n_2$.  Hence, the equation (1) becomes,

    $\frac{p_1}{T_1}=\frac{p_2}{T_2}$        (2)

The rearranged equation to calculate the final temperature of the given sample is given below.

    $T_2=\frac{p_2\times T_1}{p_1}$        (3)

Substitute the values of initial pressure, final pressure and initial temperature in equation (3).

    $\begin{array}{c}{T}_2=\frac{300\text{atm}\times 300\text{K}}{100\text{K}}\\ =\underset{\_}{900K}\end{array}$

Thus, the final temperature for the given sample of nitrogen gas is $\underset{\_}{900K}$.