Question

An ideal gas is enclosed in a cylinder with a movable piston on top of it. The piston has a mass of 7 650 g and an area of 5.10 cm2 and is free to slide up and down, keeping the pressure of the gas constant. How much work is done on the gas as the temperature of 0.180 mol of the gas is raised from 18.5°C to 340°C?

Answer 1

Step-by-step explanation:

According to the ideal gas equation, PV = nRT.

So, V =
`(nRT)/(P)` ......... (1)

Since, it is given that volume of the gas is increases. So, change in volume will be as follows.

`\Delta V = V_(f) - V_(i)`

Hence, equation (1) will become as follows.

`\Delta V` =
`(nRT_(f))/(P) - (nRT_(i))/(P)`

`P \Delta V` = nR
`(T_(f) - T_(i))`

Therefore, work done on the gas will be given as follows.

W = - P
`\Delta V`

=
`-n * R(T_(f) - T_(i))`

=
`- 0.0180 mol * 8.31 atm L/mol K (340^(o)C - 18.5^(o)C)`

= 48.09 J

Thus, we can conclude that work done on the gas is 48.09 J.