Question

Two moles of ideal He gas are contained at a pressure of 1 atm and a temperature of 300 K. 34166 J of heat are transferred to the gas, as a result of which the gas expands and does 1216 J of work against its surroundings. The process is reversible. (Note: C = 1.5R) Please calculate the final temperature of the gas

Answer 1

Step-by-step explanation:

The given data is as follows.

n = 2 mol, P = 1 atm, T = 300 K

Q = +34166 J, W= -1216 J (work done against surrounding)

`C_(v)` =
`(3R)/(2)`

Relation between internal energy, work and heat is as follows.

Change in internal energy (
`\Delta U`) = Q + W

= [34166 + (-1216)] J

= 32950 J

Also,
`\Delta U = n * C_(v) * \Delta T`

=
`3R * (T_(2) - T_(1))`

32950 J =
`3 * 8.314 J/mol K * (T_(2) - 300 K)`

`(32950)/(24.942) = T_(2) - 300 K`

1321.06 K + 300 K =
`T_(2)`

`T_(2)` = 1621.06 K

Thus, we can conclude that the final temperature of the gas is 1621.06 K.