Chapter 5, Problem 5.95P

(a)

Interpretation Introduction

Interpretation:

The volume of a gas increase, decrease or remain unchanged if the pressure is decreased from $2\text{ atm}$ to $1\text{ atm}$ while the temperature is decreased from $200°\text{C}$ to $100°\text{C}$ is to be identified.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

$PV=\text{constant}$

Here, $P$ is the pressure and $V$ is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

$V\alpha T$

Here, $T$ is the temperature and $V$ is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

$\text{V }\alpha n$

Here, $n$ is the mole of the gas and $V$ is the volume.

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

Answer to Problem 5.95P

The volume of the gas is increased if the pressure is decreased from $2\text{ atm}$ to $1\text{ atm}$ and the temperature is decreased from $200°\text{C}$ to $100°\text{C}$.

Explanation of Solution

The formula to convert $°\text{C}$ to Kelvin is:

$T_{\text{K}}=T_{\text{°C}}+273.15$ (1)

Substitute $\text{200 }°\text{C}$ for $T_{\text{°C}}$ to calculate $T_1$ in equation (1).

$\begin{array}{c}{T}_{\text{K}}=200+273.15\\ =473.15\text{ K}\end{array}$

Substitute $\text{100 }°\text{C}$ for $T_{\text{°C}}$ to calculate $T_2$ in equation (1).

$\begin{array}{c}{T}_{\text{K}}=100+273.15\\ =373.15\text{ K}\end{array}$

The expression to calculate the final volume is as follows:

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$ (2)

Here, $P_1$ is the initial pressure, $P_2$ is the final pressure, $V_1$ is the initial volume, $V_2$ is the final volume, $T_1$ is the initial temperature and $T_2$ is the final temperature.

Rearrange the equation (2) to calculate $V_2$.

$V_2=\left(\frac{P_1{V}_1{T}_2}{T_1{P}_2}\right)$ (3)

Substitute the value $\text{2 atm}$ for $P_1$, $1\text{ atm}$ for $P_2$, $373.15\text{ K}$ for $T_2$ and $\text{473}\text{.15 K}$ for $T_1$ in the equation (3)

$\begin{array}{c}{V}_2=\left(\frac{\left(V_1\right)\left(\text{373}\text{.15 K}\right)\left(\text{2 atm}\right)}{\left(\text{473}\text{.15 K}\right)\left(1\text{ atm}\right)}\right)\\ =1.577V_1\end{array}$

Hence, the volume of the gas is increased if the pressure is decreased from $2\text{ atm}$ to $1\text{ atm}$ and the temperature is decreased from $200°\text{C}$ to $100°\text{C}$.

Conclusion

The volume of the gas is increased if the pressure is decreased from $2\text{ atm}$ to $1\text{ atm}$ and the temperature is decreased from $200°\text{C}$ to $100°\text{C}$.

(b)

Interpretation Introduction

Interpretation:

The volume of a gas increase, decrease or remain unchanged if the pressure is decreased from $\text{1 atm}$ to $\text{3 atm}$ while the temperature is decreased from $100°\text{C}$ to $300°\text{C}$ is to be identified.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

$PV=\text{constant}$

Here, $P$ is the pressure and $V$ is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

$V\alpha T$

Here, $T$ is the temperature and $V$ is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

$\text{V }\alpha n$

Here, $n$ is the mole of the gas and $V$ is the volume.

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

Answer to Problem 5.95P

The volume of the gas is decreased if the pressure is decreased from $\text{1 atm}$ to $\text{3 atm}$ and the temperature is decreased from $100°\text{C}$ to $300°\text{C}$.

Explanation of Solution

Substitute $\text{100 }°\text{C}$ for $T_{\text{°C}}$ to calculate $T_1$ in equation (1).

$\begin{array}{c}{T}_{\text{K}}=200+273.15\\ =373.15\text{ K}\end{array}$

Substitute $\text{300 }°\text{C}$ for $T_{\text{°C}}$ to calculate $T_2$ in equation (1).

$\begin{array}{c}{T}_{\text{K}}=300+273.15\\ =573.15\text{ K}\end{array}$

Substitute the value $\text{1 atm}$ for $P_1$, $\text{3 atm}$ for $P_2$, $573.15\text{ K}$ for $T_2$ and $\text{373}\text{.15 K}$ for $T_1$ in the equation (3)

$\begin{array}{c}{V}_2=\left(\frac{\left(V_1\right)\left(\text{573}\text{.15 K}\right)\left(\text{1 atm}\right)}{\left(\text{373}\text{.15 K}\right)\left(\text{3 atm}\right)}\right)\\ =0.5120V_1\end{array}$

Hence, the volume of the gas is decreased if the pressure is decreased from $\text{1 atm}$ to $\text{3 atm}$ and the temperature is decreased from $100°\text{C}$ to $300°\text{C}$.

Conclusion

The volume of the gas is decreased if the pressure is decreased from $\text{1 atm}$ to $\text{3 atm}$ and the temperature is decreased from $100°\text{C}$ to $300°\text{C}$.

(c)

Interpretation Introduction

Interpretation:

The volume of a gas increase, decrease or remain unchanged if the pressure is decreased from $\text{3 atm}$ to $\text{6 atm}$ while the temperature is decreased from $-73°\text{C}$ to $\text{127 }°\text{C}$ is to be identified.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

$PV=\text{constant}$

Here, $P$ is the pressure and $V$ is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

$V\alpha T$

Here, $T$ is the temperature and $V$ is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

$\text{V }\alpha n$

Here, $n$ is the mole of the gas and $V$ is the volume.

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

Answer to Problem 5.95P

The volume of the gas is unchanged if the pressure is decreased from $\text{3 atm}$ to $\text{6 atm}$ and the temperature is decreased from $-73°\text{C}$ to $\text{127 }°\text{C}$.

Explanation of Solution

Substitute $\text{73 }°\text{C}$ for $T_{\text{°C}}$ to calculate $T_1$ in equation (1).

$\begin{array}{c}{T}_{\text{K}}=73+273.15\\ =200.15\text{ K}\end{array}$

Substitute $\text{127 }°\text{C}$ for $T_{\text{°C}}$ to calculate $T_2$ in equation (1).

$\begin{array}{c}{T}_{\text{K}}=127+273.15\\ =400.15\text{ K}\end{array}$

Substitute the value $\text{3 atm}$ for $P_1$, $\text{6 atm}$ for $P_2$, $400.15\text{ K}$ for $T_2$ and $\text{200}\text{.15 K}$ for $T_1$ in the equation (3)

$\begin{array}{c}{V}_2=\left(\frac{\left(V_1\right)\left(\text{400}\text{.15 K}\right)\left(\text{3 atm}\right)}{\left(\text{200}\text{.15 K}\right)\left(\text{6 atm}\right)}\right)\\ =0.9996V_1\end{array}$

Hence, the volume of the gas is unchanged if the pressure is decreased from $\text{3 atm}$ to $\text{6 atm}$ and the temperature is decreased from $-73°\text{C}$ to $\text{127 }°\text{C}$.

Conclusion

The volume of the gas is unchanged if the pressure is decreased from $\text{3 atm}$ to $\text{6 atm}$ and the temperature is decreased from $-73°\text{C}$ to $\text{127 }°\text{C}$.

(d)

Interpretation Introduction

Interpretation:

The volume of a gas increase, decrease or remain unchanged if the pressure is decreased from $\text{0}\text{.2 atm}$ to $\text{0}\text{.4 atm}$ while the temperature is decreased from $300°\text{C}$ to $150°\text{C}$ is to be identified.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

$PV=\text{constant}$

Here, $P$ is the pressure and $V$ is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

$V\alpha T$

Here, $T$ is the temperature and $V$ is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

$\text{V }\alpha n$

Here, $n$ is the mole of the gas and $V$ is the volume.

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

Answer to Problem 5.95P

The volume of the gas is decreased if the pressure is decreased from $\text{0}\text{.2 atm}$ to $\text{0}\text{.4 atm}$ and the temperature is decreased from $300°\text{C}$ to $150°\text{C}$.

Explanation of Solution

Substitute $\text{300 }°\text{C}$ for $T_{\text{°C}}$ to calculate $T_1$ in equation (1).

$\begin{array}{c}{T}_{\text{K}}=300+273.15\\ =573.15\text{ K}\end{array}$

Substitute $\text{150 }°\text{C}$ for $T_{\text{°C}}$ to calculate $T_2$ in equation (1).

$\begin{array}{c}{T}_{\text{K}}=150+273.15\\ =423.15\text{ K}\end{array}$

Substitute the value $\text{0}\text{.2 atm}$ for $P_1$, $\text{0}\text{.4 atm}$ for $P_2$, $423.15\text{ K}$ for $T_2$ and $\text{573}\text{.15 K}$ for $T_1$ in the equation (3)

$\begin{array}{c}{V}_2=\left(\frac{\left(V_1\right)\left(\text{423}\text{.15 K}\right)\left(\text{0}\text{.2 atm}\right)}{\left(\text{573}\text{.15 K}\right)\left(\text{0}\text{.4 atm}\right)}\right)\\ =0.369V_1\end{array}$

Hence, the volume of the gas is decreased if the pressure is decreased from $\text{0}\text{.2 atm}$ to $\text{0}\text{.4 atm}$ and the temperature is decreased from $300°\text{C}$ to $150°\text{C}$.

Conclusion

The volume of the gas is decreased if the pressure is decreased from $\text{0}\text{.2 atm}$ to $\text{0}\text{.4 atm}$ and the temperature is decreased from $300°\text{C}$ to $150°\text{C}$.