Chapter 15, Problem 8P

(a)

To determine

The work done by the gas.

Answer to Problem 8P

The work done by the gas is $78.8\text{J}$ .

Explanation of Solution

Given info:

The pressure of the gas is, $P=3\text{atm}=3\text{atm}\times \frac{1.01\times {10}^5\text{N}/{\text{m}}^2}{1\text{atm}}=3.03\times {10}^5\text{N}/{\text{m}}^2$ .

The initial volume of the gas is, $V_1=400\text{mL}=400\text{mL}\times \frac{{10}^{-6}{\text{m}}^3}{1\text{mL}}=4\times {10}^{-4}{\text{m}}^3$ .

The final volume of the gas is, $V_2=660\text{mL}=660\text{mL}\times \frac{{10}^{-6}{\text{m}}^3}{1\text{mL}}=6.6\times {10}^{-4}{\text{m}}^3$ .

Formula Used:

The expression to calculate the work done by the gas is,

  $W=P\left(V_2-V_1\right)$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}W=\left(3.03\times {10}^5\text{N}/{\text{m}}^2\right)\left(6.6\times {10}^{-4}{\text{m}}^3-4\times {10}^{-4}{\text{m}}^3\right)\\ =78.8\text{J}\end{array}$

Conclusion:

Thus, the work done by the gas is $78.8\text{J}$ .

(b)

To determine

The heat flow into the gas.

Answer to Problem 8P

The heat flow into the gas is $78.8\text{J}$ .

Explanation of Solution

Given info:

The change in the internal energy of the gas is, $\Delta U=0\text{J}$ .

Formula Used:

The expression to calculate the heat flow into the gas is,

  $Q=\Delta U+W$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}Q=0\text{J}+78.8\text{J}\\ =78.8\text{J}\end{array}$

Conclusion:

Thus, the heat flow into the gas is $78.8\text{J}$ .