Question

A piston having 7.23 g of steam at 110°C increases its temperature by 35°C. At the same time it expands from a volume of 2.00 L to 8.00 L against a constant external pressure of 0.985 bar. The heat capacity of stearm is 1.996 J g-1 Kl. Calculate q, w. Δυ, and ΔΗ in Joules. W-

Answer 1

Answer : The value of
`q,w,\Delta U\text{ and }\Delta U` is 505 J, -599 J, -94 J and -693 J respectively.

Explanation : Given,

Mass of steam = 7.23 g

Initial temperature =
`110^oC`

Final temperature =
`(110+35)^oC=145^oC`

Initial volume = 2 L

Final volume = 8 L

External pressure = 0.985 bar

Heat capacity of steam = 1.996 J/g.K

First law of thermodynamic : It states that the energy can not be created or destroyed, it can only change or transfer from one state to another state.

As per first law of thermodynamic,

`\Delta U=q+w`

First we have to calculate the heat absorbed by the system.

Formula used :

`Q=m* c* \Delta T`

or,

`Q=m* c* (T_2-T_1)`

where,

Q = heat absorbed by the system = ?

m = mass of steam = 7.23 g

`C_p` = heat capacity of steam =
`1.966J/g.K`

`T_1` = initial temperature =
`110^oC=273+110=383K`

`T_2` = final temperature =
`145^oC=273+145=418K`

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

`Q=7.23g* 1.966J/g.K* (418-383)K`

`Q=505J`

Now we have to calculate the work done.

Formula used :

`w=-p_(ext)dV\\\\w=-p_(ext)(V_2-V_1)`

where,

w = work done = ?

`p_(ext)` = external pressure = 0.985 bar = 0.985 atm (1 bar = 1 atm)

`V_1` = initial volume of gas = 2.00 L

`V_2` = final volume of gas = 8.00 L

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

`w=-p_(ext)(V_2-V_1)`

`w=-(0.985atm)* (8.00-2.00)L`

`w=-5.91L.atm=-5.91* 101.3J=-599J`

conversion used : (1 L.atm = 101.3 J)

Now we have to calculate the change in internal energy of the system.

`\Delta U=q+w`

`\Delta U=505J+(-599J)`

`\Delta U=-94J`

Now we have to calculate the change in enthalpy of the system.

Formula used :

`\Delta H=\Delta U+P\Delta V`

`\Delta H=\Delta U+w`

`\Delta H=(-94J)+(-599J)`

`\Delta H=-693J`

Therefore, the value of
`q,w,\Delta U\text{ and }\Delta U` is 505 J, -599 J, -94 J and -693 J respectively.