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Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is
(Given that 1 L bar = 100 J)
(Given that 1 L bar = 100 J)
- a)–30 J
- b)5 kJ
- c)25 J
- d)30 J
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 ...
∴ Wirr = – Pext ΔV
= – 2 bar × (0.25 – 0.1) L
= – 2 × 0.15 L-bar
= – 0.30 L-bar
= – 0.30 × 100 J
= – 30 J
= – 2 bar × (0.25 – 0.1) L
= – 2 × 0.15 L-bar
= – 0.30 L-bar
= – 0.30 × 100 J
= – 30 J
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Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 ...
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Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 ...
The work done by the gas can be calculated using the formula:
W = nRT ln(V2/V1)
where:
W = work done by the gas
n = number of moles of gas
R = gas constant (8.31 J/mol K)
T = temperature in Kelvin
V1 = initial volume
V2 = final volume
Since the gas is expanding isothermally, the temperature remains constant at 300 K. We can assume that the gas behaves ideally and use the ideal gas law to find the number of moles of gas:
PV = nRT
where:
P = external pressure
V = volume
n, R, and T are as defined above
Solving for n:
n = PV/RT
Initial conditions:
P = 2 bar
V = 0.1 L = 0.1 x 10^-3 m^3
T = 300 K
R = 8.31 J/mol K
n = (2 x 0.1 x 10^-3) / (8.31 x 300) = 8.05 x 10^-6 mol
Final conditions:
P = 2 bar
V = 0.25 L = 0.25 x 10^-3 m^3
T = 300 K
R = 8.31 J/mol K
Now we can calculate the work done by the gas:
W = nRT ln(V2/V1)
W = (8.05 x 10^-6) x (8.31) x (300) x ln(0.25/0.1)
W = 46.9 J
Finally, we can convert the answer to the desired units:
W = 46.9 J / (1 L bar / 100 J)
W = 469 L bar
W = nRT ln(V2/V1)
where:
W = work done by the gas
n = number of moles of gas
R = gas constant (8.31 J/mol K)
T = temperature in Kelvin
V1 = initial volume
V2 = final volume
Since the gas is expanding isothermally, the temperature remains constant at 300 K. We can assume that the gas behaves ideally and use the ideal gas law to find the number of moles of gas:
PV = nRT
where:
P = external pressure
V = volume
n, R, and T are as defined above
Solving for n:
n = PV/RT
Initial conditions:
P = 2 bar
V = 0.1 L = 0.1 x 10^-3 m^3
T = 300 K
R = 8.31 J/mol K
n = (2 x 0.1 x 10^-3) / (8.31 x 300) = 8.05 x 10^-6 mol
Final conditions:
P = 2 bar
V = 0.25 L = 0.25 x 10^-3 m^3
T = 300 K
R = 8.31 J/mol K
Now we can calculate the work done by the gas:
W = nRT ln(V2/V1)
W = (8.05 x 10^-6) x (8.31) x (300) x ln(0.25/0.1)
W = 46.9 J
Finally, we can convert the answer to the desired units:
W = 46.9 J / (1 L bar / 100 J)
W = 469 L bar
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Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is(Given that 1 L bar = 100 J)a)–30 Jb)5 kJc)25 Jd)30 JCorrect answer is option 'A'. Can you explain this answer?
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Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is(Given that 1 L bar = 100 J)a)–30 Jb)5 kJc)25 Jd)30 JCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is(Given that 1 L bar = 100 J)a)–30 Jb)5 kJc)25 Jd)30 JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is(Given that 1 L bar = 100 J)a)–30 Jb)5 kJc)25 Jd)30 JCorrect answer is option 'A'. Can you explain this answer?.
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