Question

A closed system contains 30g of gas. How much heat(in joules) is added to or rejected by the system to produce 5000 N-m of work while the stored energy decreases by 200 J/g?

Answer 1

Answer : The heat rejected by the system is 1000 J

Explanation :

As per first law of thermodynamic,

`q=\Delta U+w`

where,

`\Delta U` = internal energy of the system

q = heat added or rejected by the system

w = work done of the system

First we have to determine the internal energy for 30 grams of gas.

As, 1 gram of gas has internal energy = 200 J

So, 30 grams of gas has internal energy = 200 × 30 = 6000 J

Now we have to determine the heat of the system.

`q=\Delta U+w`

`\Delta` = -6000 J

w = 5000 N.m = 5000 J

Now put all the given values in the above formula, we get:

`q=-6000J+5000J`

`q=-1000J`

The negative sign indicate that the heat rejected by the system.

Hence, the heat rejected by the system is 1000 J