Advertisements
Advertisements
प्रश्न
Answer the following question.
Calculate the maximum work when 24 g of O2 are expanded isothermally and reversibly from the pressure of 1.6 bar to 1 bar at 298 K.
उत्तर
Given:
Mass of O2 = 24 g
Initial pressure = P1 = 1.6 bar
Final pressure = P2 = 1 bar
Temperature = T = 298 K
To find: Maximum work (Wmax)
Formula: `"W"_"max" = - 2.303 "nRT" log_10 "P"_1/"P"_2`
Calculation:
Number of moles of O2 = n = `(24 "g")/(32 "g mol"^-1)` = 0.75 mol
Gas constant = R = 8.314 J K–1 mol–1
Now, using formula,
`"W"_"max" = - 2.303 "nRT" log_10 "P"_1/"P"_2`
= - 2.303 × 0.75 mol × 8.314 J K-1mol-1 × 298 K × `log_10 1.6/1`
= - 2.303 × 0.75 × 8.314 J × 298 × 0.2041
= - 873.4 J
The maximum work done is – 873.4 J.
APPEARS IN
संबंधित प्रश्न
Answer the following question.
Derive the expression for the maximum work.
Write the expression to calculate maximum work done when 1 mole of an ideal gas expands isothermally and reversibly from V1 to V2.
0.022 kg of CO2 is compressed isothermally and reversibly at 298 K from an initial pressure of 100 kPa when the work obtained is 1200 J, calculate the final pressure.
An ideal gas expands from 1 × 10−3 m3 to 1 × 10−2 m3 at 300 K against a constant external pressure of 1 × 105 N m−2, work done is ____________.
What is the work done when 2 mole of an ideal gas is expanded isothermally and reversibly from 5 m3 to 10 m3 at 300 K? (R = 8.314 J K-1 mol-1)
One mole of an ideal gas is expanded isothermally and reversibly from 10 L to 15 L at 300 K. Calculate the work done in the process.
2000 mmol of an ideal gas expanded isothermally and reversibly from 20 L to 30 L at 300 K, calculate the work done in the process (R = 8.314 JK–1 mol–1).
Calculate work done when 0.5 moles of C02 is compressed isothermally and reversibly at 298 K from 100 kPa to 263.4 kPa. (R = 8.314 kJ mol-1 K-1)
Three moles of ideal gas is compressed isothermally and reversibly to a volume of 2 dm3 . The work done is 2.984 kJ at 22°c. Calculate V1 for the gas.
