Chapter 15, Problem 2P

(a)

To determine

To find: Work done by Gas.

Answer to Problem 2P

  $6.3\times {10}^5$ Joule

Explanation of Solution

Given Data:

  $V_f=18.2m^3$

  $V_i=12m^3$

  $P=1$ atm

Formula used: $workdone=P\times \Delta V$

Calculation:

Work done $=10V$

  $\Rightarrow w={10}^5\times \left(18.2-12\right)$

  $\Rightarrow w=6.262\times {10}^5$ Joule

Conclusion:

Hence work done will be $6.3\times {10}^5$ Joule.

(b)

To determine

To find: Change in internal energy.

Answer to Problem 2P

The change in internal energy is $5.2\times {10}^{6}J$ .

Explanation of Solution

Given Data:

Heat supplied $=1400$ kcal

Formula Used:

  $\begin{array}{l}\Delta U=Q-w\\ =(1400\text{kcal})\left(\frac{4186\text{J}}{1\text{kcal}}\right)-\left(6.262\times {10}^5\text{J}\right)\\ =5.2\times {10}^{6}J\end{array}$

Hence change in internal energy will be $5.2\times {10}^{6}J$ J.