Posted by **Mary** on Sunday, April 15, 2007 at 2:04pm.

Three moles of an ideal monatomic gas are at a temperature of 345 K. Then, 2531 J of heat are added to the gas, and 1101 J of work are done on it. What is the final temperature of the gas?

delta U= 3/2nR(T final -T initial)

(2531J - 1101J) = 3/2(3.0mol)(8.31)(T final - 345K)

1430J = 37.395(T final - 345K)

1430J/ 37.395 = T final - 345K

38.24 + 345K = T final

383.2404K = T final

This answer is incorrect. Please explain to me where I went wrong.

## Answer This Question

## Related Questions

- Physics repost please check - Three moles of an ideal monatomic gas are at a ...
- Physics please check - Three moles of an ideal monatomic gas are at a ...
- Physics please check - Three moles of an ideal monatomic gas are at a ...
- Physics - Three moles of an ideal monatomic gas are at a temperature of 396 K. ...
- Physics - Three moles of an ideal monatomic gas are at a temperature of 396 K. ...
- Physics - Three moles of an ideal monoatomic gas are at a temperature of 345 K. ...
- Phyics - Three moles of an ideal monatomic gas are at a temperature of 396 K. ...
- Physics - A tank with a constant volume of 3.44 m3 contains 14 moles of a ...
- Physics - A cylinder with a movable piston contains 14 moles of a monatomic ...
- help me please physics - A monatomic ideal gas expands from point A to point B ...

More Related Questions