Question

) Two thousand Joules of energy are added to a 5 l container of gaseous helium atoms at 300 K. Determine the final temperature of the gas. Then consider adding the same amount of energy to the same container with N2 gas, and then with propane gas. Which of the three will reach the highest temperature? The lowest temperature? Why?

Answer 1

Step-by-step explanation:

The given three gases are helium, nitrogen and propane.

It is known that
`C_(p)` values of each of these gases is as follows.

`C_(p)(nitrogen)` = 1.044 (KJ/Kg.K)

`C_(p)(helium)` = 5.19 (KJ/Kg.K)

`C_(p)(propane)` = 1.67 (KJ/Kg.K)

Now, relation between heat and
`C_(p)` is as follows.

Q =
`n_(helium) * C_(p)(helium) * [T_(final) - 300]`

2 =
`(\rho_(helium) * V_(helium)) * C_(p)(helium) * [T_(final) - 300]`

2 =
`(0.179 * 0.005) * 5.19 [T_(final) - 300]`

`T_(final)` = 730.57 K

In the same way, Q =
`n_(nitrogen) * C_(p)(nitrogen) * [T_(final) - 300]`

2 =
`(1.2506 * 0.005) * 1.044 [T_(final) - 300]`

`T_(final)` = 606.37 K

For propane also, Q =
`n_(propane) * C_(p)(propane) * [T_(final) - 300]`

2 =
`(1.4506 * 0.005) * 1.67 [T_(final) - 300]`

`T_(final)` = 465.12 K

Here, we can see that there occurs least increase in temperature of propane whereas helium showed more increase in temperature.

This change arises because there density of helium is very less as compared to the density of propane and nitrogen. As a result, there occurs more turbulence in case of helium because of which it has less density and more rise in temperature.