Chapter 13, Problem 91GP

(a)

To determine

To Find: The volume of air for the given conditions.

Answer to Problem 91GP

The volume of air at $1\text{ atm}$ pressure and $20°\text{C}$ temperature is $2204\text{ L}$ .

Explanation of Solution

Given:

Pressure of air at state 1, $P_1=195\text{ atm}$

Pressure of air at state 2, $P_2=1\text{ atm}$

Volume of air at state 1, $V_1=11.3\text{ L}$

Temperature of air at state 1 and state 2, $T_1=T_2={20}^{\circ }\text{C}$

Formula used:

Ideal gas law formula is used,

  $\begin{array}{l}P\times V=n\times R\times T\\ \frac{P\times V}{T}=n\times R\text{ = Constant}\end{array}$

Here, $P$ is the pressure of the air,

  $V$ is the volume of the air,

  $n$ is the number of molecules,

  $R$ is the universal gas constant,

  $T$ is the Temperature of the air.

Calculation:

The above equation for different pressure and volume is rewritten as follows and then substitute the given values in the above equation,

  $\begin{array}{l}\frac{P_1\times V_1}{T_1}=\frac{P_2\times V_2}{T_2}\\ \text{But, }{T}_1=T_2\\ P_1\times V_1=P_2\times V_2\\ V_2=\frac{P_1\times V_1}{P_2}=\frac{195\text{ atm}\times 11.3}{1\text{ atm}}\text{ = }2204\text{ L}\end{array}$

Conclusion:

Thus, the volume of air at $1\text{ atm}$ pressure and ${20}^{\circ }C$ temperature is $2204\text{ L}$ .

(b)

To determine

To Find: The time for emptying the tank.

Answer to Problem 91GP

The time for emptying the tank is $92\text{ s}$ .

Explanation of Solution

Given:

Person consumes air in each breath $\text{ = 2 L}$ in each breath.

Number of breaths per minute $\text{= }12\text{ breath / minute}$ .

Total air consumed $\Delta V=2\times 12=24\text{ L / minute}$ .

Total volume of air $V=2204\text{ L}$ .

Formula used:

Use the above information to find the time required for emptying the tank from the following formula:

Time required to empty the tank, $t=\frac{V}{\Delta V}$ .

Here, $\Delta V$ is the total air consumed,

  $V$ is the total volume of air.

Calculation:

Substitute the given values in the above equation:

Time required to empty the tank $t=\frac{2204}{24}=91.8\text{ s}$ .

Time upto which the tank last is $92\text{s}$ .

Conclusion:

Thus, the time up to which the tanklast is $92\text{ s}$ .

(c)

To determine

To Find: The time for emptying the tankat a depth of $20\text{ m}$ .

Answer to Problem 91GP

The time for emptying the tankat a depth of $20\text{ m}$ of sea water and at a temperature of $10°\text{C}$ is $30.16\text{ min}$ .

Explanation of Solution

Given:

Pressure $P_1=195\text{ atm = }1.97\times {10}^7\text{ Pa}$ .

Pressure at depth $20\text{ m, }{P}_2=1\text{ atm + }\left(\rho \times g\times d\right)$ .

Volume $V_1=11.3\text{ L}$ .

Temperature $T_1={20}^{\circ }\text{C = }293\text{ K}$ .

Depth $d=20\text{ m}$ .

Temperature $T_2={10}^{\circ }\text{C = }283\text{ K}$ .

Rate of flow $Q=24\text{ L / min}$ .

Formula used:

Ideal gas law formula is used,

  $\begin{array}{l}P\times V=n\times R\times T\\ \frac{P\times V}{T}=n\times R\text{ = Constant}\\ Q=\frac{V}{t}\end{array}$

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}\frac{P_1\times V_1}{T_1}=\frac{P_2\times V_2}{T_2}\\ V_2=\frac{P_1\times V_1}{T_1}\times \frac{T_2}{P_2}=\frac{1.97\times {10}^7\times 11.3\times 283}{293\times 2.97\times {10}^5}=724\text{ L}\\ Q\text{ = }\frac{\text{Volume}}{\text{Time}}\\ t\text{ = }\frac{\text{Volume}}{\text{Rate of flow}}=\frac{724}{24}=30.16\text{ min}\end{array}$

Conclusion:

Thus, the time for emptying the tankat a depth of 20 m of sea water and at a temperature of ${10}^{\circ }C$ is $30.16\text{ min}$ .