Chapter 15, Problem 10P

(a)

To determine

The work done by the gas.

Answer to Problem 10P

The work done by the gas is $352.5\text{J}$ .

Explanation of Solution

Given info:

The initial pressure of the gas is, $P_1=2.2\text{atm}=2.2\text{atm}\times \frac{1.01\times {10}^5\text{N}/{\text{m}}^2}{1\text{atm}}=2.22\times {10}^5\text{N}/{\text{m}}^2$ .

The final pressure of the gas is, $P_2=1.4\text{atm}=1.4\text{atm}\times \frac{1.01\times {10}^5\text{N}/{\text{m}}^2}{1\text{atm}}=1.41\times {10}^5\text{N}/{\text{m}}^2$ .

The initial volume of the gas is, $V_1=6.8\text{L}=6.8\text{L}\times \frac{{10}^{-3}{\text{m}}^3}{1\text{L}}=6.8\times {10}^{-3}{\text{m}}^3$ .

The final volume of the gas is, $V_2=9.3\text{L}=9.3\text{L}\times \frac{{10}^{-3}{\text{m}}^3}{1\text{L}}=9.3\times {10}^{-3}{\text{m}}^3$ .

Formula Used:

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

  $W=P_2\left(V_2-V_1\right)$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}W=\left(1.41\times {10}^5\text{N}/{\text{m}}^2\right)\left(9.3\times {10}^{-3}{\text{m}}^3-6.8\times {10}^{-3}{\text{m}}^3\right)\\ =352.5\text{J}\end{array}$

Conclusion:

Thus, the work done by the gas is $352.5\text{J}$ .

(b)

To determine

The change in the internal energy of the gas.

Answer to Problem 10P

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

Explanation of Solution

Given info:

The change in the temperature of the gas is, $\Delta T=0°\text{C}$ .

Formula Used:

The expression to calculate the change in the internal energy of the gas is,

  $\Delta U=\frac{1}{2}N\cdot R\cdot \Delta T$

Here, N is the number of moles, R is the gas constant

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}\Delta U=\frac{1}{2}NR\left(0°\text{C}\right)\\ =0\text{J}\end{array}$

Conclusion:

Thus, the change in the internal energy of the gas is $0\text{J}$ .

(c)

To determine

The heat flow into the gas.

Answer to Problem 10P

The heat flow into the gas is $352.5\text{J}$ .

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.

  $\begin{array}{c}Q=0\text{J}+352.5\text{J}\\ =352.5\text{J}\end{array}$

Conclusion:

Thus, the heat flow into the gas is $352.5\text{J}$ .