A 3.25 mol sample of an ideal diatomic gas expandsadiabatically from a volume of

Question

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

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.