One litres of an ideal gas at 10 atmosphere expands isothermally and r...
Given information:
- Initial volume (V1) = 1 liter
- Final volume (V2) = 10 liters
- Pressure (P) = 10 atm
- Isothermal process
Formula:
The work done during an isothermal expansion of an ideal gas is given by the equation:
W = nRT ln(V2/V1)
Where:
W = work done
n = number of moles of the gas
R = ideal gas constant
T = temperature
ln = natural logarithm
Solution:
Step 1: Determine the number of moles of the gas
To find the number of moles (n), we can use the ideal gas equation: PV = nRT.
Given:
P = 10 atm
V = 1 liter
R = ideal gas constant
Rearranging the equation, we get:
n = PV / RT
Substituting the given values:
n = (10 atm) * (1 liter) / (0.0821 L.atm/mol.K * T)
Since the temperature (T) is not given, we cannot calculate the exact value of 'n'. However, we can assume a value for T and proceed with the calculation.
Step 2: Calculate the work done using the formula
W = nRT ln(V2/V1)
Given:
V1 = 1 liter
V2 = 10 liters
Substituting the values, we get:
W = nRT ln(10/1)
Step 3: Simplify the expression
Using logarithmic properties, ln(a/b) = ln(a) - ln(b), we can simplify the expression:
W = nRT (ln(10) - ln(1))
W = nRT ln(10)
Step 4: Final calculation
The final calculation depends on the assumed value of temperature (T) in Step 1. Without the exact value of T, we cannot determine the specific work done. However, we can evaluate the answer options provided using the formula W = nRT ln(10).
Answer options:
(A) -10L atm
(B) -23.03L atm
(C) -46.06L atm
(D) -101.3L atm
Without the specific value of temperature, we cannot determine which answer option is correct. The correct answer will depend on the actual value of T.
One litres of an ideal gas at 10 atmosphere expands isothermally and r...
Ans is 0 bcoz work in vacuum is zero I think or D
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