A 3.25 mol sample of an ideal diatomic gas expandsadiabatically from a volume of 0.1990 m^3 to 0.704 m^3.Initially the pressure was 1.00 atm.. Initial temp 746K and Final temp 450K change in internal energy is -2.00*10^4 J heat loss is )J Determine the work done on the gas. (Assume nomolecular vibration.) A 3.25 mol sample of an ideal diatomic gas expandsadiabatically from a volume of 0.1990 m^3 to 0.704 m^3.Initially the pressure was 1.00 atm.. Initial temp 746K and Final temp 450K change in internal energy is -2.00*10^4 J heat loss is )J Determine the work done on the gas. (Assume nomolecular vibration.)
Explanation / Answer
first law of thermo: change ininternal energy = heat added or lost + work done on the gas . adiabatic means heat lost or added is zero so... . work done on the gas = changein internal energy = -2.00 x 104J