Question

describe the first law of thermodynamics. consider a thermodynamic process, which move the system from state - 1 to state - 2 by doing 152 j amount of work and internal energy of the system increase by 68 j. (a) find the total heat energy added or lose? (b) if the system then moves from state - 2 back to state - 1, what is its change in internal energy? (c) when moving along a different path between same 2 states 278 j of work is done on the system, how much heat does it absorb or release? (a) q = δe+w=(152 + 68)j =? j. i, w = + 152 j, δe = 68 j = e₂ - e₁ (b) ii, δe = e_f - e_i = e₁ - e₂ = δe = 1/2 nrδt δe = e₁ - e₂ = -(e₂ - e₁)= - 68 j. (c) q = δe+w : along # iii 1→2 q=(+68 - 278)j q =? j. w=-278 j (on) n→constant isolated. 68 j = δe₁→₂ = 1/2 nr(t₂ - t₁) - 68 j = δe₂→₁ = 1/2 nr(t₁ - t₂) - 1/2 nr(t₂ - t₁) 68

Explanation:

Step1: State the first - law of thermodynamics

The first law of thermodynamics is $\Delta U = Q - W$, where $\Delta U$ is the change in internal energy of the system, $Q$ is the heat added to the system, and $W$ is the work done by the system.

Step2: Solve part (a)

Given $W = 152J$ (work done by the system) and $\Delta U=68J$. Using $\Delta U = Q - W$, we can find $Q$. Rearranging the formula gives $Q=\Delta U + W$. Substituting the values, $Q=(152 + 68)J=220J$.

Step3: Solve part (b)

When the system moves from state - 2 back to state - 1, the change in internal energy $\Delta U$ is the negative of the change in internal energy when moving from state - 1 to state - 2. So $\Delta U=- 68J$.

Step4: Solve part (c)

Given $\Delta U = 68J$ (since the change in internal energy between two states is path - independent) and $W=-278J$ (work done on the system). Using $\Delta U = Q - W$, we can find $Q$. Rearranging gives $Q=\Delta U+W$. Substituting values, $Q=(68-278)J=-210J$. The negative sign indicates that heat is released by the system.

Answer:

(a) $220J$
(b) $-68J$
(c) $-210J$