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Find the maximum work done (cal) when 0.5 mol of a gas expands isothermally and reversibly from a volume of 2L to 5L at 27°C and calculate the change in internal energy if 200 cal of heat is absrobed? [rounded up to two decimal places]
Correct answer is between '102.00,104.00'. Can you explain this answer?
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Find the maximum work done (cal) when 0.5 mol of a gas expands isother...
°C.
First, we need to determine the pressure of the gas using the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
We are given that n = 0.5 mol, V1 = 2 L, V2 = 5 L, and T = 27°C = 300 K (since we need to use Kelvin in the ideal gas law).
Rearranging the equation, we get:
P = nRT/V
Plugging in the values, we get:
P = (0.5 mol)(0.08206 L·atm/mol·K)(300 K)/(2 L) = 6.123 atm
Now, we can use the formula for work done during an isothermal expansion:
W = -nRT ln(V2/V1)
where W is the work done, n is the number of moles, R is the gas constant, T is the temperature in Kelvin, and ln is the natural logarithm.
Plugging in the values, we get:
W = -(0.5 mol)(0.08206 L·atm/mol·K)(300 K) ln(5/2) = -40.8 cal
Note that the negative sign indicates that work is being done on the system (since the gas is expanding), and the work done is measured in calories (cal).
Therefore, the maximum work done is 40.8 cal.
First, we need to determine the pressure of the gas using the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
We are given that n = 0.5 mol, V1 = 2 L, V2 = 5 L, and T = 27°C = 300 K (since we need to use Kelvin in the ideal gas law).
Rearranging the equation, we get:
P = nRT/V
Plugging in the values, we get:
P = (0.5 mol)(0.08206 L·atm/mol·K)(300 K)/(2 L) = 6.123 atm
Now, we can use the formula for work done during an isothermal expansion:
W = -nRT ln(V2/V1)
where W is the work done, n is the number of moles, R is the gas constant, T is the temperature in Kelvin, and ln is the natural logarithm.
Plugging in the values, we get:
W = -(0.5 mol)(0.08206 L·atm/mol·K)(300 K) ln(5/2) = -40.8 cal
Note that the negative sign indicates that work is being done on the system (since the gas is expanding), and the work done is measured in calories (cal).
Therefore, the maximum work done is 40.8 cal.
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Find the maximum work done (cal) when 0.5 mol of a gas expands isothermally and reversibly from a volume of 2L to 5L at 27°C and calculate the change in internal energy if 200 cal of heat is absrobed? [rounded up to two decimal places]Correct answer is between '102.00,104.00'. Can you explain this answer?
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Find the maximum work done (cal) when 0.5 mol of a gas expands isothermally and reversibly from a volume of 2L to 5L at 27°C and calculate the change in internal energy if 200 cal of heat is absrobed? [rounded up to two decimal places]Correct answer is between '102.00,104.00'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Find the maximum work done (cal) when 0.5 mol of a gas expands isothermally and reversibly from a volume of 2L to 5L at 27°C and calculate the change in internal energy if 200 cal of heat is absrobed? [rounded up to two decimal places]Correct answer is between '102.00,104.00'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find the maximum work done (cal) when 0.5 mol of a gas expands isothermally and reversibly from a volume of 2L to 5L at 27°C and calculate the change in internal energy if 200 cal of heat is absrobed? [rounded up to two decimal places]Correct answer is between '102.00,104.00'. Can you explain this answer?.
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