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If 22 g of CO2, showing ideal behaviour, isothermally and reversibly from 2.5 L to 7.5 L at 27°C, then calculate the work done (J) and heat absorbed: [rounded up to two decimal places]
Correct answer is between '1368.00,1372.00'. Can you explain this answer?
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If 22 g of CO2, showing ideal behaviour, isothermally and reversibly f...
°C, what is the work done by the gas?
We can use the formula for the work done by an ideal gas during an isothermal and reversible process:
W = nRT ln(V2/V1)
where n is the number of moles of gas, R is the gas constant, T is the temperature in Kelvin, V1 and V2 are the initial and final volumes, respectively.
First, we need to calculate the number of moles of CO2:
n = m/M
where m is the mass of CO2 and M is the molar mass of CO2.
M = 12.01 + 2(15.99) = 44.01 g/mol
n = 22 g / 44.01 g/mol = 0.4995 mol
Next, we need to convert the temperature to Kelvin:
T = 27 + 273.15 = 300.15 K
Now we can plug in the values and solve for W:
W = (0.4995 mol)(0.0821 L·atm/mol·K)(300.15 K) ln(7.5 L/2.5 L)
W = 12.13 atm·L
Therefore, the work done by the gas is 12.13 atm·L.
We can use the formula for the work done by an ideal gas during an isothermal and reversible process:
W = nRT ln(V2/V1)
where n is the number of moles of gas, R is the gas constant, T is the temperature in Kelvin, V1 and V2 are the initial and final volumes, respectively.
First, we need to calculate the number of moles of CO2:
n = m/M
where m is the mass of CO2 and M is the molar mass of CO2.
M = 12.01 + 2(15.99) = 44.01 g/mol
n = 22 g / 44.01 g/mol = 0.4995 mol
Next, we need to convert the temperature to Kelvin:
T = 27 + 273.15 = 300.15 K
Now we can plug in the values and solve for W:
W = (0.4995 mol)(0.0821 L·atm/mol·K)(300.15 K) ln(7.5 L/2.5 L)
W = 12.13 atm·L
Therefore, the work done by the gas is 12.13 atm·L.
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Community Answer
If 22 g of CO2, showing ideal behaviour, isothermally and reversibly f...
Thus, the work done by the gas is approximately -1370.45 Joules and the heat absorbed is approximately 1370.45 Joules
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