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6 litres of an ideal gas expands isothermally at a temperature of 300 Kelvin up to 10 litres at a pressure of 5 atm, what is the work done?
- a)30 Newton metre
- b)80 Newton metre
- c)50 Newton metres
- d)20 Newton metre
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
6 litres of an ideal gas expands isothermally at a temperature of 300 ...
Given:
Initial volume of gas (V1) = 6 litres
Final volume of gas (V2) = 10 litres
Temperature (T) = 300 Kelvin
Pressure (P) = 5 atm
To find: Work done (W)
The work done by an ideal gas during an isothermal expansion can be calculated using the formula:
W = nRT ln(V2/V1)
Where:
n = number of moles of gas
R = ideal gas constant (8.314 J/mol·K)
ln = natural logarithm
V1 = initial volume of gas
V2 = final volume of gas
To find the number of moles (n), we can use the ideal gas equation:
PV = nRT
Rearranging the equation, we have:
n = PV / RT
Substituting the given values:
n = (5 atm * 6 liters) / (0.0821 atm·L/mol·K * 300 K)
n ≈ 0.98 moles
Substituting the values of n, R, V1, and V2 into the formula for work done:
W = (0.98 mol * 8.314 J/mol·K * 300 K) ln(10 L / 6 L)
W ≈ 19.6 J ln(10/6)
W ≈ 19.6 J ln(5/3)
W ≈ 19.6 J ln(1.67)
W ≈ 19.6 J * 0.51
W ≈ 9.996 J
W ≈ 10 J
Therefore, the work done by the gas during the isothermal expansion is approximately 10 Joules.
Initial volume of gas (V1) = 6 litres
Final volume of gas (V2) = 10 litres
Temperature (T) = 300 Kelvin
Pressure (P) = 5 atm
To find: Work done (W)
The work done by an ideal gas during an isothermal expansion can be calculated using the formula:
W = nRT ln(V2/V1)
Where:
n = number of moles of gas
R = ideal gas constant (8.314 J/mol·K)
ln = natural logarithm
V1 = initial volume of gas
V2 = final volume of gas
To find the number of moles (n), we can use the ideal gas equation:
PV = nRT
Rearranging the equation, we have:
n = PV / RT
Substituting the given values:
n = (5 atm * 6 liters) / (0.0821 atm·L/mol·K * 300 K)
n ≈ 0.98 moles
Substituting the values of n, R, V1, and V2 into the formula for work done:
W = (0.98 mol * 8.314 J/mol·K * 300 K) ln(10 L / 6 L)
W ≈ 19.6 J ln(10/6)
W ≈ 19.6 J ln(5/3)
W ≈ 19.6 J ln(1.67)
W ≈ 19.6 J * 0.51
W ≈ 9.996 J
W ≈ 10 J
Therefore, the work done by the gas during the isothermal expansion is approximately 10 Joules.
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Community Answer
6 litres of an ideal gas expands isothermally at a temperature of 300 ...
The expression for work done is given by pressure x volume difference, here an ideal gas has a volume difference of 4 litres at 5 ATM pressure. So the work done = 10 – 6 liters x 5atm = 20 Newton metre.
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6 litres of an ideal gas expands isothermally at a temperature of 300 Kelvin up to 10 litres at a pressure of 5 atm, what is the work done?a)30 Newton metreb)80 Newton metrec)50 Newton metresd)20 Newton metreCorrect answer is option 'D'. Can you explain this answer?
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