Solveeit Logo

Question

Question: The \(28{\text{ g}}\) of \({{\text{N}}_2}\) gas at \(300{\text{ K}}\) and \(20{\text{ atm}}\) was al...

The 28 g28{\text{ g}} of N2{{\text{N}}_2} gas at 300 K300{\text{ K}} and 20 atm20{\text{ atm}} was allowed to expand isothermally against a constant external pressure of 1 atm1{\text{ atm}}. ΔU\Delta {\text{U}}, q{\text{q}} and w{\text{w}} for the gas is:
A. ΔU=0, q=2370 J, w=23.70×102 J\Delta {\text{U}} = 0,{\text{ q}} = 2370{\text{ J, w}} = - 23.70 \times {10^2}{\text{ J}}
B. ΔU=2000, q=2371 J, w=24.70×102 J\Delta {\text{U}} = 2000,{\text{ q}} = 2371{\text{ J, w}} = 24.70 \times {10^2}{\text{ J}}
C. ΔU=0, q=2374 J, w=23.70×102 J\Delta {\text{U}} = 0,{\text{ q}} = 2374{\text{ J, w}} = - 23.70 \times {10^2}{\text{ J}}
D. ΔU=0, q=2370 J, w=24.70×102 J\Delta {\text{U}} = 0,{\text{ q}} = 2370{\text{ J, w}} = - 24.70 \times {10^2}{\text{ J}}

Explanation

Solution

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 (ΔU)\left( {\Delta {\text{U}}} \right), heat (q)\left( {\text{q}} \right) and work (w)\left( {\text{w}} \right).

Formulae Used:
1. w=PexternalΔV{\text{w}} = - {{\text{P}}_{{\text{external}}}}\Delta {\text{V}}
2. Number of moles(mol)=Mass(g)Molar mass(g/mol){\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}
3. ΔU=q+w\Delta {\text{U}} = {\text{q}} + {\text{w}}

Complete answer:
1. Calculate the ΔU\Delta {\text{U}} for the gas:
We are given that 28 g28{\text{ g}} of N2{{\text{N}}_2} gas at 300 K300{\text{ K}} and 20 atm20{\text{ atm}} was allowed to expand isothermally against a constant external pressure of 1 atm1{\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 (U)\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 (ΔU)\left( {\Delta {\text{U}}} \right) is zero. Thus, ΔU=0\Delta {\text{U}} = 0.

2. Calculate the w{\text{w}} for the gas:
We know that the expression for the work when a gas expands isothermally is,
w=Pexternal(V2V1){\text{w}} = - {{\text{P}}_{{\text{external}}}}\left( {{{\text{V}}_2} - {{\text{V}}_1}} \right) …… (1)
where
w{\text{w}} is the work,
Pexternal{{\text{P}}_{{\text{external}}}} is the external pressure,
V2{{\text{V}}_2} is the final volume,
V1{{\text{V}}_1} is the initial volume.
Calculate the number of moles of N2{{\text{N}}_2} gas in 28 g28{\text{ g}} of N2{{\text{N}}_2} gas as follows:
Number of moles(mol)=Mass(g)Molar mass(g/mol){\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 g28{\text{ g}} for the mass of gas, 28 g/mol28{\text{ g/mol}} for the molar mass of gas. Thus,
Number of moles=28 g28 g/mol{\text{Number of moles}} = \dfrac{{28{\text{ g}}}}{{28{\text{ g/mol}}}}
Number of moles=1 mol{\text{Number of moles}} = 1{\text{ mol}}
Thus, the number of moles of N2{{\text{N}}_2} gas in 28 g28{\text{ g}} of N2{{\text{N}}_2} gas are 1 mol1{\text{ mol}}.
Calculate the initial volume of the gas using the expression for ideal gas as follows:
P1V1=nRT{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}} = {\text{nRT}}
Where,
P1{{\text{P}}_{\text{1}}} is the initial pressure of the gas,
V1{{\text{V}}_{\text{1}}} is the initial volume of the gas,
n{\text{n}} is the number of moles of gas,
R{\text{R}} is the universal gas constant,
T{\text{T}} is the temperature of the gas.
Substitute 20 atm20{\text{ atm}} for the initial pressure of the gas, 1 mol1{\text{ mol}} for the number of moles of gas, 300 K300{\text{ K}} for the temperature of the gas. Thus,
V1=nRTP1{{\text{V}}_{\text{1}}} = \dfrac{{{\text{nRT}}}}{{{{\text{P}}_{\text{1}}}}}
V1=1 mol×R×300 K20 atm\Rightarrow {{\text{V}}_{\text{1}}} = \dfrac{{1{\text{ mol}} \times {\text{R}} \times 300{\text{ K}}}}{{20{\text{ atm}}}}
V1=15R\Rightarrow {{\text{V}}_{\text{1}}} = 15{\text{R}}
Thus, the initial volume of the gas is 15R15{\text{R}}.
Similarly, calculate the final volume of the gas using the expression for ideal gas as follows:
P2V2=nRT{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}} = {\text{nRT}}
where,
P2{{\text{P}}_2} is the final pressure of the gas,
V2{{\text{V}}_2} is the final volume of the gas,
n{\text{n}} is the number of moles of gas,
R{\text{R}} is the universal gas constant,
T{\text{T}} is the temperature of the gas.
Substitute 1 atm1{\text{ atm}} for the final pressure of the gas, 1 mol1{\text{ mol}} for the number of moles of gas, 300 K300{\text{ K}} for the temperature of the gas. Thus,
V2=nRTP2{{\text{V}}_{\text{2}}} = \dfrac{{{\text{nRT}}}}{{{{\text{P}}_{\text{2}}}}}
V2=1 mol×R×300 K1 atm\Rightarrow {{\text{V}}_{\text{2}}} = \dfrac{{1{\text{ mol}} \times {\text{R}} \times 300{\text{ K}}}}{{1{\text{ atm}}}}
V2=300R\Rightarrow {{\text{V}}_{\text{2}}} = 300{\text{R}}
Thus, the final volume of the gas is 300R300{\text{R}}.
Now, calculate the work for the has using equation (1) as follows:
w=Pexternal(V2V1){\text{w}} = - {{\text{P}}_{{\text{external}}}}\left( {{{\text{V}}_2} - {{\text{V}}_1}} \right) …… (1)
Substitute 1 atm1{\text{ atm}} for the external pressure, 300R300{\text{R}} for the final volume and 15R15{\text{R}} for the initial volume. Thus,
w=1 atm×(300R15R)\Rightarrow {\text{w}} = - 1{\text{ atm}} \times \left( {300{\text{R}} - 15{\text{R}}} \right)
w=1 atm×(285R)\Rightarrow {\text{w}} = - 1{\text{ atm}} \times \left( {285{\text{R}}} \right)
Substitute 8.314 J mol1 K18.314{\text{ J mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}} for the universal gas constant. Thus,
w=1 atm×(285×8.314 J mol1 K1)\Rightarrow {\text{w}} = - 1{\text{ atm}} \times \left( {285 \times 8.314{\text{ J mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}} \right)
w=2370 J\Rightarrow {\text{w}} = - 2370{\text{ J}}
Thus, w=23.70×102 J{\text{w}} = - 23.70 \times {10^2}{\text{ J}}

3. Calculate the q{\text{q}} for the gas:
We know the expression for first law of thermodynamics is,
ΔU=q+w\Delta {\text{U}} = {\text{q}} + {\text{w}}
where
ΔU\Delta {\text{U}} is the change in internal energy,
q{\text{q}} is heat,
w{\text{w}} is work.
We know that ΔU=0\Delta {\text{U}} = 0. Thus,
q=w{\text{q}} = - {\text{w}}
Substitute w=23.70×102 J{\text{w}} = - 23.70 \times {10^2}{\text{ J}}. Thus,
q=(23.70×102 J)\Rightarrow {\text{q}} = - \left( { - 23.70 \times {{10}^2}{\text{ J}}} \right)
q=23.70×102 J\Rightarrow {\text{q}} = 23.70 \times {10^2}{\text{ J}}
Thus, q=2370 J{\text{q}} = 2370{\text{ J}}
Thus, ΔU=0, q=2370 J, w=23.70×102 J\Delta {\text{U}} = 0,{\text{ q}} = 2370{\text{ J, w}} = - 23.70 \times {10^2}{\text{ J}}

**Thus, the correct option is (A) ΔU=0, q=2370 J, w=23.70×102 J\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.