Chapter 15, Problem 13P

(a)

To determine

The change in the internal energy.

Answer to Problem 13P

The change in the internal energy is 25 Joule.

Explanation of Solution

Given info:

Energy released from the system, $Q_{ac}=-80\text{J}$

Work done by the system, $W_{ac}=55\text{J}$

Formula Used:

Write the expression to calculate the change in the internal energy.

  $\begin{array}{c}{U}_{\text{change}}=U_a-U_c\\ =-Q_{ac}-W_{ac}\end{array}$.......... (1)

Where,

• $U_{\text{change}}$is the change in the internal energy.

Calculation:

Substitute the values in equation (1).

  $\begin{array}{c}{U}_{\text{change}}=-\left(-80\right)-\left(55\right)\\ =80-55\\ =25\text{J}\end{array}$

Hence, the change in the internal energy is 25 Joule.

Conclusion:

Therefore, the change in the internal energy is 25 Joule.

(b)

To determine

The amount of heat added in the process cda.

Answer to Problem 13P

The amount of heat added in the process is $63\text{J}$ .

Explanation of Solution

Given info:

Work done during the process, $W_{\text{cda}}=38\text{J}$

Formula Used:

Write the expression to calculate amount of heat added.

  $\Delta U_{\text{cda}}=U_{\text{change}}+W_{\text{cda}}$.......... (2)

Where,

• $U_{\text{cda}}$is the amount of heat added.

Calculation:

Substitute the values in equation (2).

  $\begin{array}{c}\Delta U_{\text{cda}}=25+38\\ =63\text{J}\end{array}$

Hence, the amount of heat added in the process is $63\text{J}$ .

Conclusion:

Therefore, the amount of heat added in the process is $63\text{J}$ .

(c)

To determine

The work done by the gas in the process abc.

Answer to Problem 13P

The work done by the gas in the process $-95\text{J}$ .

Explanation of Solution

Given info:

Relation between pressure, $P_a=2.5P_d$

Consider the given cycle.

  

Formula Used:

Write the expression to calculate work done by gas in the process abc.

  $W_{\text{abc}}=P_a\Delta V_{ab}$.......... (2)

Where,

• $W_{\text{abc}}$is the work done by the gas.
• $P_a$is the pressure of the gas.
• $\Delta V_{ab}$is the change in the volume.

Calculation:

Since, the pressure of the gas at a is 2.5 times the pressure of the gas at d. So,

  $W_{\text{abc}}=2.5P_d\left(V_c-V_d\right)$

The work done by the gas in the process cda is equal to $P_d\left(V_c-V_d\right)$ . Therefore,

  $W_{\text{abc}}=-2.5\left(W_{\text{cda}}\right)$

Substitute the values in the above expression.

  $\begin{array}{c}{W}_{\text{abc}}=-2.5\left(38\right)\\ =-95\text{J}\end{array}$

Conclusion:

Therefore, the work done by the gas in the process $-95\text{J}$ .

(d)

To determine

The amount of the heat $Q$ in the process abc.

Answer to Problem 13P

The amount of the heat $Q$ in the process abcis $-120\text{J}$ .

Explanation of Solution

Given info:

Consider the given cycle.

  

Formula Used:

Write the expression to calculate the heat in the process abc.

  $Q_{\text{abc}}=-\Delta U_{\text{abc}}+\left(-W_{abc}\right)$

Calculation:

Substitute the values in the above equation.

  $\begin{array}{c}{Q}_{\text{abc}}=-25-95\\ =-120\text{J}\end{array}$

Conclusion:

Therefore, the amount of the heat $Q$ in the process abcis $-120\text{J}$ .

(e)

To determine

The amount of the heat $Q$ for the process bcif $U_a-U_b=10\text{J}$ .

Answer to Problem 13P

The amount of the heat $Q$ for the process bcis $-15\text{J}$ .

Explanation of Solution

Given info:

Consider the given cycle.

  

Formula Used:

Write the expression to calculate the change in the internal energy during the process bc.

  $\Delta U_{\text{bc}}=U_{\text{c}}-U_{\text{b}}$

Calculation:

Consider the given relation.

  $\begin{array}{c}{U}_a-U_b=10\text{J}\\ U_b=U_a-10\end{array}$

Substitute the value in the above expression.

  $\begin{array}{c}\Delta U_{\text{bc}}=U_{\text{c}}-\left(U_a-10\right)\\ =U_{\text{c}}-U_a+10\\ =\Delta U_{\text{ca}}+10\end{array}$

Substitute the value in the above expression.

  $\begin{array}{c}\Delta U_{\text{bc}}=-25+10\\ =-15\text{J}\end{array}$

Conclusion:

Therefore, the amount of the heat $Q$ for the process bcis $-15\text{J}$ .