Chapter 13, Problem 34P

(a)

To determine

To Calculate: The volume of the helium gas by using the ideal gas equation.

Answer to Problem 34P

The volume of the helium gas is $\text{0}\text{.323}{\text{m}}^{\text{3}}$ .

Explanation of Solution

Given:

Number of moles, $n=18.75\text{ mol}$

Temperature, $T=10{}^{\text{o}}\text{C}$

Pressure, $\mathrm{P}=0.350\text{ atm}$

Formula used:

The ideal gas equation for the gas,

Where $P=$ Pressure of the gas

  $V=$ The volume of the gas

  $R=$ Gas constant

  $T=$ The temperature of the gas

  $\begin{array}{l}PV=nRT\\ V=\frac{nRT}{P}\end{array}$

Calculation:

The absolute pressure is given by:

  $P=1\text{ atm + 0}\text{.350 atm = 1}\text{.350 atm}$

Consider that helium gas is the ideal gas. So, the volume of the helium gas can be derived using the ideal gas equation

  $\begin{array}{c}PV=nRT\\ V=\frac{nRT}{P}\\ =\frac{\text{(18}\text{.75}\text{mol)(8}\text{.315}\text{J/mol}\cdot \text{K)(283}\text{K)}}{\text{(1}\text{.350}\text{atm)(1}{\text{.013×10}}^{\text{5}}\text{Pa/atm)}}\\ \text{=0}\text{.323}{\text{m}}^{\text{3}}\end{array}$

Conclusion:

The volume of the helium gas is $\text{0}\text{.323}{\text{m}}^{\text{3}}$ .

(b)

To determine

To Calculate: The temperature of the gas as per according to the information provided.

Answer to Problem 34P

The temperature change is $\text{210}\text{K}$ .

Explanation of Solution

Given:

  $\begin{array}{l}{T}_1=283\text{K}\\ P_1=1\text{ atm + 0}\text{.350 atm = 1}\text{.350 atm}\end{array}$

Formula used:

The ideal gas equation for the gas,

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

Where $P=$ Pressure of the gas.

  $V=$ The volume of the gas

  $R=$ Gas constant

  $T=$ The temperature of the gas

Calculation:

For given amount of gas, $\frac{PV}{T}=\text{constant}$

Also, the pressure:

  $P_2=1\text{ atm + 1 atm = 2 atm}$

Also, according to the statement,

  $V_2=\frac{1}{2}{V}_1$

So,

  $\begin{array}{l}\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}\\ T_2=T_1\frac{P_2{V}_2}{P_1{V}_1}\\ =T_1\left(\frac{P_2}{P_1}\right)\left(\frac{\frac{1}{2}{V}_1}{V_1}\right)\\ =(\text{283}\text{K)}\left(\frac{\text{2}\text{atm}}{\text{1}\text{.350}\text{atm}}\right)\left(\frac{1}{2}\right)\\ =\text{210}\text{K}\end{array}$

Conclusion:

The amount of gas is remains unchanged and the temperature change is $\text{210}\text{K}$ .