Question

A stationary cylinder of oxygen used in a hospital has the following characteristics at room temperature 300k, gauge pressure$1 \cdot 38 \times {10^7}{\text{Pa}}$, volume\[16L\]. If the flow area, measured at atmospheric pressure, is constant $2 \cdot 4\dfrac{L}{{\min .}}$ the cylinder last nearly for,
A) 5h
B) 10h
C) 15h
D) 20h

Answer 1

Hint:The pressure of the cylinder after the cylinder gets empty is equal to the atmospheric pressure as the cylinder is getting empty at room temperature. The volume of the cylinder is the flow of the cylinder in the time taken by the cylinder to get empty.

Formula used: The formula for the isothermal process is given by ${P_1}{V_1} = {P_2}{V_2}$ where ${P_1}$ is the initial pressure of the cylinder ${V_1}$ is the initial volume of the cylinder ${P_2}$ is the final pressure of the cylinder and ${V_2}$is the final volume of the cylinder.

Step by step solution:
It is given that the initial pressure is equal to $1 \cdot 38 \times {10^7}{\text{Pa}}$ also the initial volume is \[16L\] also the
Final pressure will be atmospheric pressure of${P_2} = {10^5}Pa$ and final volume will be equal to ${V_2} = 2 \cdot 4t$ where t is the time taken to empty the cylinder.
In the isothermal process we get,
$ \Rightarrow {P_1}{V_1} = {P_2}{V_2}$
Replace the value of${P_1}$,${P_2}$ and ${V_1}$ we get
$ \Rightarrow \left( {1 \cdot 38 \times {{10}^7}} \right)\left( {16} \right) = \left( {{{10}^5}} \right){V_2}$
\[ \Rightarrow 2 \cdot 4t = \dfrac{{\left( {1 \cdot 38 \times {{10}^7}} \right)\left( {16} \right)}}{{\left( {{{10}^5}} \right)}}\]
\[ \Rightarrow 2 \cdot 4t = \left( {1 \cdot 38 \times {{10}^2}} \right)\left( {16} \right)\]
\[ \Rightarrow 2 \cdot 4t = 2208\]
\[ \Rightarrow t = \dfrac{{2208}}{{2 \cdot 4}}\]
\[ \Rightarrow t = 920\min .\]
Converting minutes into hours.
\[ \Rightarrow t = \dfrac{{920}}{{60}}h\]
\[ \Rightarrow t = 15.3h\]
\[ \Rightarrow t \simeq 15h\]
So the time taken by the cylinder to empty itself is\[t = 15h\].

The correct option for this problem is option C.

Additional information:The isothermal temperature means that the temperature is always constant. The work done for isothermal process is given by $W = {P_1}{V_1}\ln \left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)$ where W is the work done ${P_1}$ is the initial pressure ${V_1}$ is the initial volume of the cylinder and ${V_2}$ is the final volume of the cylinder.

Note:The relation for the isothermal process is based on the property of constant temperature. The ideal gas equation is the main equation from which the isothermal process relation of the body is taken out by keeping temperature as constant.