Posted by **AC** on Thursday, June 21, 2007 at 11:33am.

Suppose 2500 J of heat are added to 4.1 mol of argon gas at a constant pressure of 120 kPa. (Assume that the argon can be treated as an ideal monatomic gas.) Find the change in internal energy.

I started using U = N (3/2 kB T). But I have too many unknowns. The only other equation that our teacher gave us is Won + Qon = deltaU (Work on system + Heat on system = Change in internal energy). Please help!

Qon = deltaU - Won =

deltaU + Wsystem,

where Wsystem is the work performed by the system (in this case the argon gas). Because the pressure is constant, the work performed is:

P deltaV = Delta (PV) = (using ideal gas law) Delta (N kB T)

This means that:

Qon = deltaU + Wsystem =

5/2 N kB delta T =

(5/2)(2/3)(3/2)N kB delta T =

(5/2)(2/3) delta U = 5/3 delta U

And it follows that:

delta U = 3/5 Qon = 1500 J

## Answer this Question

## Related Questions

- Physics - Suppose 2100 J of heat are added to 3.3 mol of argon gas at a constant...
- physics - A cylinder with a movable piston holds 2.75 mol of argon gas at a ...
- Physics - A cylinder with a moveable piston holds 2.60 mol of argon at a ...
- Physics I - A container of volume 0.71 m3 contains 1 mol of argon gas at 27&#...
- college phyics - A system consists of 3.2 mol of the monatomic gas neon (which ...
- physics - A cylinder with a moveable piston holds 3.00 of argon at a constant ...
- chemistry - A sealed container comtains 1.50 mol of nitrogen gas, 2.00 mol of ...
- Chemistry - A sealed container comtains 1.50 mol of nitrogen gas, 2.00 mol of ...
- physics - A container of volume 0.39 m3 contains 8.7 mol of argon gas at 32&#...
- Physics - 0.75 mol of argon gas is admitted to an evacuated 40 cm^{3} container ...