The $28{\text{ g}}$ of ${{\text{N}}_2}$ gas at $300{\text{ K}}$ and $20{\text{ atm}}$ was allowed to expand isothermally against a constant external pressure of $1{\text{ atm}}$. $\Delta {\text{U}}$, ${\text{q}}$ and ${\text{w}}$ for the gas is:
A. $\Delta {\text{U}} = 0,{\text{ q}} = 2370{\text{ J, w}} = - 23.70 \times {10^2}{\text{ J}}$
B. $\Delta {\text{U}} = 2000,{\text{ q}} = 2371{\text{ J, w}} = 24.70 \times {10^2}{\text{ J}}$
C. $\Delta {\text{U}} = 0,{\text{ q}} = 2374{\text{ J, w}} = - 23.70 \times {10^2}{\text{ J}}$
D. $\Delta {\text{U}} = 0,{\text{ q}} = 2370{\text{ J, w}} = - 24.70 \times {10^2}{\text{ J}}$
Answer
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Hint:To solve this we must know the expression for the first law of thermodynamics. The expression gives the relationship between the change in internal energy $\left( {\Delta {\text{U}}} \right)$, heat $\left( {\text{q}} \right)$ and work $\left( {\text{w}} \right)$.
Formulae Used:
1. ${\text{w}} = - {{\text{P}}_{{\text{external}}}}\Delta {\text{V}}$
2. ${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
3. $\Delta {\text{U}} = {\text{q}} + {\text{w}}$
Complete answer:
1. Calculate the $\Delta {\text{U}}$ for the gas:
We are given that $28{\text{ g}}$ of ${{\text{N}}_2}$ gas at $300{\text{ K}}$ and $20{\text{ atm}}$ was allowed to expand isothermally against a constant external pressure of $1{\text{ atm}}$.
We know that in an isothermal process, the temperature of the system remains constant or unchanged throughout the process.
For an ideal gas, the internal energy $\left( {\text{U}} \right)$ depends on the temperature. At a constant temperature (isothermal condition), the internal energy of the gas is constant. Thus, the change in internal energy $\left( {\Delta {\text{U}}} \right)$ is zero. Thus, $\Delta {\text{U}} = 0$.
2. Calculate the ${\text{w}}$ for the gas:
We know that the expression for the work when a gas expands isothermally is,
${\text{w}} = - {{\text{P}}_{{\text{external}}}}\left( {{{\text{V}}_2} - {{\text{V}}_1}} \right)$ …… (1)
where
${\text{w}}$ is the work,
${{\text{P}}_{{\text{external}}}}$ is the external pressure,
${{\text{V}}_2}$ is the final volume,
${{\text{V}}_1}$ is the initial volume.
Calculate the number of moles of ${{\text{N}}_2}$ gas in $28{\text{ g}}$ of ${{\text{N}}_2}$ gas as follows:
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
Substitute $28{\text{ g}}$ for the mass of gas, $28{\text{ g/mol}}$ for the molar mass of gas. Thus,
${\text{Number of moles}} = \dfrac{{28{\text{ g}}}}{{28{\text{ g/mol}}}}$
${\text{Number of moles}} = 1{\text{ mol}}$
Thus, the number of moles of ${{\text{N}}_2}$ gas in $28{\text{ g}}$ of ${{\text{N}}_2}$ gas are $1{\text{ mol}}$.
Calculate the initial volume of the gas using the expression for ideal gas as follows:
${{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}} = {\text{nRT}}$
Where,
${{\text{P}}_{\text{1}}}$ is the initial pressure of the gas,
${{\text{V}}_{\text{1}}}$ is the initial volume of the gas,
${\text{n}}$ is the number of moles of gas,
${\text{R}}$ is the universal gas constant,
${\text{T}}$ is the temperature of the gas.
Substitute $20{\text{ atm}}$ for the initial pressure of the gas, $1{\text{ mol}}$ for the number of moles of gas, $300{\text{ K}}$ for the temperature of the gas. Thus,
${{\text{V}}_{\text{1}}} = \dfrac{{{\text{nRT}}}}{{{{\text{P}}_{\text{1}}}}}$
$\Rightarrow {{\text{V}}_{\text{1}}} = \dfrac{{1{\text{ mol}} \times {\text{R}} \times 300{\text{ K}}}}{{20{\text{ atm}}}}$
$\Rightarrow {{\text{V}}_{\text{1}}} = 15{\text{R}}$
Thus, the initial volume of the gas is $15{\text{R}}$.
Similarly, calculate the final volume of the gas using the expression for ideal gas as follows:
${{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}} = {\text{nRT}}$
where,
${{\text{P}}_2}$ is the final pressure of the gas,
${{\text{V}}_2}$ is the final volume of the gas,
${\text{n}}$ is the number of moles of gas,
${\text{R}}$ is the universal gas constant,
${\text{T}}$ is the temperature of the gas.
Substitute $1{\text{ atm}}$ for the final pressure of the gas, $1{\text{ mol}}$ for the number of moles of gas, $300{\text{ K}}$ for the temperature of the gas. Thus,
${{\text{V}}_{\text{2}}} = \dfrac{{{\text{nRT}}}}{{{{\text{P}}_{\text{2}}}}}$
$\Rightarrow {{\text{V}}_{\text{2}}} = \dfrac{{1{\text{ mol}} \times {\text{R}} \times 300{\text{ K}}}}{{1{\text{ atm}}}}$
$\Rightarrow {{\text{V}}_{\text{2}}} = 300{\text{R}}$
Thus, the final volume of the gas is $300{\text{R}}$.
Now, calculate the work for the has using equation (1) as follows:
${\text{w}} = - {{\text{P}}_{{\text{external}}}}\left( {{{\text{V}}_2} - {{\text{V}}_1}} \right)$ …… (1)
Substitute $1{\text{ atm}}$ for the external pressure, $300{\text{R}}$ for the final volume and $15{\text{R}}$ for the initial volume. Thus,
$\Rightarrow {\text{w}} = - 1{\text{ atm}} \times \left( {300{\text{R}} - 15{\text{R}}} \right)$
$\Rightarrow {\text{w}} = - 1{\text{ atm}} \times \left( {285{\text{R}}} \right)$
Substitute $8.314{\text{ J mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}$ for the universal gas constant. Thus,
$\Rightarrow {\text{w}} = - 1{\text{ atm}} \times \left( {285 \times 8.314{\text{ J mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}} \right)$
$\Rightarrow {\text{w}} = - 2370{\text{ J}}$
Thus, ${\text{w}} = - 23.70 \times {10^2}{\text{ J}}$
3. Calculate the ${\text{q}}$ for the gas:
We know the expression for first law of thermodynamics is,
$\Delta {\text{U}} = {\text{q}} + {\text{w}}$
where
$\Delta {\text{U}}$ is the change in internal energy,
${\text{q}}$ is heat,
${\text{w}}$ is work.
We know that $\Delta {\text{U}} = 0$. Thus,
${\text{q}} = - {\text{w}}$
Substitute ${\text{w}} = - 23.70 \times {10^2}{\text{ J}}$. Thus,
$\Rightarrow {\text{q}} = - \left( { - 23.70 \times {{10}^2}{\text{ J}}} \right)$
$\Rightarrow {\text{q}} = 23.70 \times {10^2}{\text{ J}}$
Thus, ${\text{q}} = 2370{\text{ J}}$
Thus, $\Delta {\text{U}} = 0,{\text{ q}} = 2370{\text{ J, w}} = - 23.70 \times {10^2}{\text{ J}}$
Thus, the correct option is (A) $\Delta {\text{U}} = 0,{\text{ q}} = 2370{\text{ J, w}} = - 23.70 \times {10^2}{\text{ J}}$.
Note:
The gas expands isothermally under external pressure. Thus, we use the expression for work done for irreversible isothermal expansion of gas. The changes are irreversible because external pressure is applied.
Formulae Used:
1. ${\text{w}} = - {{\text{P}}_{{\text{external}}}}\Delta {\text{V}}$
2. ${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
3. $\Delta {\text{U}} = {\text{q}} + {\text{w}}$
Complete answer:
1. Calculate the $\Delta {\text{U}}$ for the gas:
We are given that $28{\text{ g}}$ of ${{\text{N}}_2}$ gas at $300{\text{ K}}$ and $20{\text{ atm}}$ was allowed to expand isothermally against a constant external pressure of $1{\text{ atm}}$.
We know that in an isothermal process, the temperature of the system remains constant or unchanged throughout the process.
For an ideal gas, the internal energy $\left( {\text{U}} \right)$ depends on the temperature. At a constant temperature (isothermal condition), the internal energy of the gas is constant. Thus, the change in internal energy $\left( {\Delta {\text{U}}} \right)$ is zero. Thus, $\Delta {\text{U}} = 0$.
2. Calculate the ${\text{w}}$ for the gas:
We know that the expression for the work when a gas expands isothermally is,
${\text{w}} = - {{\text{P}}_{{\text{external}}}}\left( {{{\text{V}}_2} - {{\text{V}}_1}} \right)$ …… (1)
where
${\text{w}}$ is the work,
${{\text{P}}_{{\text{external}}}}$ is the external pressure,
${{\text{V}}_2}$ is the final volume,
${{\text{V}}_1}$ is the initial volume.
Calculate the number of moles of ${{\text{N}}_2}$ gas in $28{\text{ g}}$ of ${{\text{N}}_2}$ gas as follows:
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
Substitute $28{\text{ g}}$ for the mass of gas, $28{\text{ g/mol}}$ for the molar mass of gas. Thus,
${\text{Number of moles}} = \dfrac{{28{\text{ g}}}}{{28{\text{ g/mol}}}}$
${\text{Number of moles}} = 1{\text{ mol}}$
Thus, the number of moles of ${{\text{N}}_2}$ gas in $28{\text{ g}}$ of ${{\text{N}}_2}$ gas are $1{\text{ mol}}$.
Calculate the initial volume of the gas using the expression for ideal gas as follows:
${{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}} = {\text{nRT}}$
Where,
${{\text{P}}_{\text{1}}}$ is the initial pressure of the gas,
${{\text{V}}_{\text{1}}}$ is the initial volume of the gas,
${\text{n}}$ is the number of moles of gas,
${\text{R}}$ is the universal gas constant,
${\text{T}}$ is the temperature of the gas.
Substitute $20{\text{ atm}}$ for the initial pressure of the gas, $1{\text{ mol}}$ for the number of moles of gas, $300{\text{ K}}$ for the temperature of the gas. Thus,
${{\text{V}}_{\text{1}}} = \dfrac{{{\text{nRT}}}}{{{{\text{P}}_{\text{1}}}}}$
$\Rightarrow {{\text{V}}_{\text{1}}} = \dfrac{{1{\text{ mol}} \times {\text{R}} \times 300{\text{ K}}}}{{20{\text{ atm}}}}$
$\Rightarrow {{\text{V}}_{\text{1}}} = 15{\text{R}}$
Thus, the initial volume of the gas is $15{\text{R}}$.
Similarly, calculate the final volume of the gas using the expression for ideal gas as follows:
${{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}} = {\text{nRT}}$
where,
${{\text{P}}_2}$ is the final pressure of the gas,
${{\text{V}}_2}$ is the final volume of the gas,
${\text{n}}$ is the number of moles of gas,
${\text{R}}$ is the universal gas constant,
${\text{T}}$ is the temperature of the gas.
Substitute $1{\text{ atm}}$ for the final pressure of the gas, $1{\text{ mol}}$ for the number of moles of gas, $300{\text{ K}}$ for the temperature of the gas. Thus,
${{\text{V}}_{\text{2}}} = \dfrac{{{\text{nRT}}}}{{{{\text{P}}_{\text{2}}}}}$
$\Rightarrow {{\text{V}}_{\text{2}}} = \dfrac{{1{\text{ mol}} \times {\text{R}} \times 300{\text{ K}}}}{{1{\text{ atm}}}}$
$\Rightarrow {{\text{V}}_{\text{2}}} = 300{\text{R}}$
Thus, the final volume of the gas is $300{\text{R}}$.
Now, calculate the work for the has using equation (1) as follows:
${\text{w}} = - {{\text{P}}_{{\text{external}}}}\left( {{{\text{V}}_2} - {{\text{V}}_1}} \right)$ …… (1)
Substitute $1{\text{ atm}}$ for the external pressure, $300{\text{R}}$ for the final volume and $15{\text{R}}$ for the initial volume. Thus,
$\Rightarrow {\text{w}} = - 1{\text{ atm}} \times \left( {300{\text{R}} - 15{\text{R}}} \right)$
$\Rightarrow {\text{w}} = - 1{\text{ atm}} \times \left( {285{\text{R}}} \right)$
Substitute $8.314{\text{ J mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}$ for the universal gas constant. Thus,
$\Rightarrow {\text{w}} = - 1{\text{ atm}} \times \left( {285 \times 8.314{\text{ J mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}} \right)$
$\Rightarrow {\text{w}} = - 2370{\text{ J}}$
Thus, ${\text{w}} = - 23.70 \times {10^2}{\text{ J}}$
3. Calculate the ${\text{q}}$ for the gas:
We know the expression for first law of thermodynamics is,
$\Delta {\text{U}} = {\text{q}} + {\text{w}}$
where
$\Delta {\text{U}}$ is the change in internal energy,
${\text{q}}$ is heat,
${\text{w}}$ is work.
We know that $\Delta {\text{U}} = 0$. Thus,
${\text{q}} = - {\text{w}}$
Substitute ${\text{w}} = - 23.70 \times {10^2}{\text{ J}}$. Thus,
$\Rightarrow {\text{q}} = - \left( { - 23.70 \times {{10}^2}{\text{ J}}} \right)$
$\Rightarrow {\text{q}} = 23.70 \times {10^2}{\text{ J}}$
Thus, ${\text{q}} = 2370{\text{ J}}$
Thus, $\Delta {\text{U}} = 0,{\text{ q}} = 2370{\text{ J, w}} = - 23.70 \times {10^2}{\text{ J}}$
Thus, the correct option is (A) $\Delta {\text{U}} = 0,{\text{ q}} = 2370{\text{ J, w}} = - 23.70 \times {10^2}{\text{ J}}$.
Note:
The gas expands isothermally under external pressure. Thus, we use the expression for work done for irreversible isothermal expansion of gas. The changes are irreversible because external pressure is applied.
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