Thermodynamic work Answers to task 2

Answers to task 2

1)      W = P·(V₂ – V₁)

W = 2·10⁵ Pa·(3-1)· 10^-3 m³

W = 400 J

2)      T = const => P₁·V₁ = P₂·V₂

As a bubble rises up hydrostatic pressure acting on it is decreasing (P₁ > P₂), so volume is increasing (V₁ < V₂, i.e. gas performs work)

3)      Yes, during isothermal expansion (T = const) all the heat transferred to the body is obtained to do work (Q = W).

4)      There is not enough space for the same number of molecules to move freely. The collisions are increasing in number (both between the molecules and with the walls of the container), i.e. increasing pressure and consequently temperature of the gas.

5)      W = P·(V₂ – V₁)

P =

P =  

P = 150 kPa

6)      Work is the area under the graph line; i.e. the bigger the area the bigger work is done. That makes it process a.

According to the formula  =  the parameters and their changes are independent on the process; so the temperature difference in both cases is the same. It means that the change in internal energy is also the same.

7)      W = P·(V₂ – V₁)

W = 500·10³ Pa·(33-7)· 10^-3 m³

W = 13 kJ