Question

A chemical system is sealed in a strong, rigid container at room temperature, and then heated vigorously.

(a) State whether$\mathbf{∆}\mathit{U}$ , q, and wof the system are positive, negative, or zero during the heating process.

(b) Next, the container is cooled to its original temperature. Determine the signs of $\mathbf{∆}\mathit{U}$, q, and wfor the cooling process.

(c) Designate heating as step 1 and cooling as step 2. Determine the signs of $\left(∆U_1+∆U_2\right)$,$\left(q_1+q_2\right)$ , and $\left(w_1+w_2\right)$ if possible.

Short answer

(a) w=0

q is positive.

The value of$∆U$ is also positive.

(b) w=0

Q is negative.

$\text{∆U = q + w}$

The value of $∆U$is negative.

(c)

$$\begin{gathered} {\text{U}}_{\text{1}}{\text{+U}}_{\text{2}}\text{= 0} \\ \text{q}{}_{\text{1}}\text{+q}_{\text{2}}\text{= 0} \\ {\text{w}}_{\text{1}}{\text{+w}}_{\text{2}}\text{= 0} \end{gathered}$$

Step-by-step solution

First law of thermodynamics

The equation for first law of thermodynamics connects the internal energy, heat and work and the equation is:
$\mathbf{∆}\mathbf{U}\mathbf{=}\mathbf{q}\mathbf{+}\mathbf{w}$

 Predicting the values of internal energy, heat and work

(a) The system is sealed and rigid.

Internal energy is directly proportional to the temperature.

So, when the temperature of the system increases, the internal energy increases.

When the temperature increases, system takes up heat from the surroundings. So q must be positive.

Volume is constant , so work w=0.

So,

w=0

q is positive.

$\text{∆U = q + w}$

The value of $∆U$is also positive.

(b) When the system is cooled, the temperature of the system decreases leading to a decrease in the internal energy. When the temperature decreases, the system gives up heat to the surroundings. So the value of q is negative.

Volume is constant, so work w=0.

So,

W=0

Q is negative.

role="math" localid="1663412505120" $\text{∆U = q + w}$

The value of $∆U$is negative.

(c) We know that from the first law of thermodynamics,

$\text{∆U = q + w}$

For both the processes,

$$\begin{gathered} {\text{U}}_{\text{1}}{\text{=q}}_{\text{1}}{\text{+w}}_{\text{1}} \\ {\text{U}}_{\text{2}}\text{= q}{}_{\text{2}}\text{+w}_{\text{2}} \\ {\text{U =U}}_{\text{1}}{\text{+U}}_{\text{2}} \\ {\text{=q}}_{\text{1}}{\text{+q}}_{\text{2}}{\text{+w}}_{\text{1}}{\text{+w}}_{\text{2}} \end{gathered}$$

The value will be zero as this is a state function and the container goes back to the original temperature (constant volume).

So,

$$\begin{gathered} {\text{U}}_{\text{1}}{\text{+U}}_{\text{2}}\text{= 0} \\ \text{q}{}_{\text{1}}\text{+q}_{\text{2}}\text{= 0} \\ {\text{w}}_{\text{1}}{\text{+w}}_{\text{2}}\text{= 0} \end{gathered}$$