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A quantity of 0.50 mole of an ideal gas at 20°C expands isothermally against a constant pressure of 2.0 atm from 1.0 L to 5.0 L. Entropy change of the system is:
- a)6.7 JK–1
- b)–2.8 JK–1
- c)3.9 JK–1
- d)None is correct
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A quantity of 0.50 mole of an ideal gas at 20°C expands isothermal...
°C and 1 atm pressure occupies a volume of approximately 11.2 L. This can be calculated using the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant (0.08206 L·atm/mol·K), and T is the temperature in Kelvin (K).
First, we need to convert the temperature from Celsius to Kelvin:
T = 20°C + 273.15 = 293.15 K
Then we can rearrange the ideal gas law to solve for volume:
V = nRT/P
V = (0.50 mol)(0.08206 L·atm/mol·K)(293.15 K)/(1 atm)
V ≈ 11.2 L
Therefore, a quantity of 0.50 mole of an ideal gas at 20°C and 1 atm pressure occupies a volume of approximately 11.2 L.
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant (0.08206 L·atm/mol·K), and T is the temperature in Kelvin (K).
First, we need to convert the temperature from Celsius to Kelvin:
T = 20°C + 273.15 = 293.15 K
Then we can rearrange the ideal gas law to solve for volume:
V = nRT/P
V = (0.50 mol)(0.08206 L·atm/mol·K)(293.15 K)/(1 atm)
V ≈ 11.2 L
Therefore, a quantity of 0.50 mole of an ideal gas at 20°C and 1 atm pressure occupies a volume of approximately 11.2 L.
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Community Answer
A quantity of 0.50 mole of an ideal gas at 20°C expands isothermal...
Del S = n[Cv*ln(T2/T1) + RTln(v2/v1)]
As we know process is isothermal,
Hence T2=T1
Del S= n[RTln(v2/v1)]
Put values get Answer !
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A quantity of 0.50 mole of an ideal gas at 20°C expands isothermally against a constant pressure of 2.0 atm from 1.0 L to 5.0 L. Entropy change of the system is:a)6.7 JK–1 b)–2.8 JK–1c)3.9 JK–1d)None is correctCorrect answer is option 'A'. Can you explain this answer?
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