Question
Question: The \(28{\text{ g}}\) of \({{\text{N}}_2}\) gas at \(300{\text{ K}}\) and \(20{\text{ atm}}\) was al...
The 28 g of N2 gas at 300 K and 20 atm was allowed to expand isothermally against a constant external pressure of 1 atm. ΔU, q and w for the gas is:
A. ΔU=0, q=2370 J, w=−23.70×102 J
B. ΔU=2000, q=2371 J, w=24.70×102 J
C. ΔU=0, q=2374 J, w=−23.70×102 J
D. ΔU=0, q=2370 J, w=−24.70×102 J
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), heat (q) and work (w).
Formulae Used:
1. w=−PexternalΔV
2. Number of moles(mol)=Molar mass(g/mol)Mass(g)
3. ΔU=q+w
Complete answer:
1. Calculate the ΔU for the gas:
We are given that 28 g of N2 gas at 300 K and 20 atm was allowed to expand isothermally against a constant external pressure of 1 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) 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) is zero. Thus, ΔU=0.
2. Calculate the w for the gas:
We know that the expression for the work when a gas expands isothermally is,
w=−Pexternal(V2−V1) …… (1)
where
w is the work,
Pexternal is the external pressure,
V2 is the final volume,
V1 is the initial volume.
Calculate the number of moles of N2 gas in 28 g of N2 gas as follows:
Number of moles(mol)=Molar mass(g/mol)Mass(g)
Substitute 28 g for the mass of gas, 28 g/mol for the molar mass of gas. Thus,
Number of moles=28 g/mol28 g
Number of moles=1 mol
Thus, the number of moles of N2 gas in 28 g of N2 gas are 1 mol.
Calculate the initial volume of the gas using the expression for ideal gas as follows:
P1V1=nRT
Where,
P1 is the initial pressure of the gas,
V1 is the initial volume of the gas,
n is the number of moles of gas,
R is the universal gas constant,
T is the temperature of the gas.
Substitute 20 atm for the initial pressure of the gas, 1 mol for the number of moles of gas, 300 K for the temperature of the gas. Thus,
V1=P1nRT
⇒V1=20 atm1 mol×R×300 K
⇒V1=15R
Thus, the initial volume of the gas is 15R.
Similarly, calculate the final volume of the gas using the expression for ideal gas as follows:
P2V2=nRT
where,
P2 is the final pressure of the gas,
V2 is the final volume of the gas,
n is the number of moles of gas,
R is the universal gas constant,
T is the temperature of the gas.
Substitute 1 atm for the final pressure of the gas, 1 mol for the number of moles of gas, 300 K for the temperature of the gas. Thus,
V2=P2nRT
⇒V2=1 atm1 mol×R×300 K
⇒V2=300R
Thus, the final volume of the gas is 300R.
Now, calculate the work for the has using equation (1) as follows:
w=−Pexternal(V2−V1) …… (1)
Substitute 1 atm for the external pressure, 300R for the final volume and 15R for the initial volume. Thus,
⇒w=−1 atm×(300R−15R)
⇒w=−1 atm×(285R)
Substitute 8.314 J mol−1 K−1 for the universal gas constant. Thus,
⇒w=−1 atm×(285×8.314 J mol−1 K−1)
⇒w=−2370 J
Thus, w=−23.70×102 J
3. Calculate the q for the gas:
We know the expression for first law of thermodynamics is,
ΔU=q+w
where
ΔU is the change in internal energy,
q is heat,
w is work.
We know that ΔU=0. Thus,
q=−w
Substitute w=−23.70×102 J. Thus,
⇒q=−(−23.70×102 J)
⇒q=23.70×102 J
Thus, q=2370 J
Thus, ΔU=0, q=2370 J, w=−23.70×102 J
**Thus, the correct option is (A) ΔU=0, q=2370 J, w=−23.70×102 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.