Question

The pressure of an oxygen containing cylinder is 10.3atm.Under same temperature, another oxygen cylinder having 1/3rd volume and has a pressure 1.1 atm Keeping the temperature same, two cylinders are connected, find the total pressure of the system​

Answer 1

Since the temperature of the system remains constant, Boyle's Law is applicable here, i.e., the product of pressure and volume remains constant.

$\sf{\longrightarrow P_iV_i=P_fV_f}$

Let the volume of the large container be V, then volume of small container will be V/3.

Pressure of gas in large container is 10.3 atm and that of gas in small container is 1.1 atm.

Initial product of pressure and volume of the system,

$\longrightarrow\sf{P_i}\sf{V_i}=\sf{10.3V}+\sf{\dfrac{1.1V}{3}}$

When the two cylinders are connected, the volume of the system becomes V + V/3 = 4V/3. Let P be the final pressure of the system.

Final product of pressure and volume of the system,

$\longrightarrow\sf{P_fV_f}=\sf{\dfrac{4PV}{3}}$

Now,

$\longrightarrow\sf{10.3V}+\sf{\dfrac{1.1V}{3}}=\sf{\dfrac{4PV}{3}}$

$\longrightarrow\sf{\dfrac{4P}{3}}=\sf{10.3}+\sf{\dfrac{1.1}{3}}$

$\longrightarrow\sf{\dfrac{4P}{3}}=\sf{\dfrac{32.0}{3}}$

$\longrightarrow\underline{\underline{\sf{P}=\sf{8.0\ atm}}}$

Answer 2

Concept Used :-

Boyle's Law

• As per Boyle’s law, any change in the volume occupied by a gas (at constant quantity and temperature) will result in a change in the pressure exerted by it.
• In other words, the product of the initial pressure and the initial volume of a gas is equal to the product of its final pressure and final volume (at constant temperature and number of moles).
• This law can be expressed mathematically as follows:

$\sf \: P_1V_1 = P_2V_2$

where,

$\rm \: P_1  \: is \: the \: initial \: pressure \: exerted \: by \: the \: gas$

$\rm \: V_1  \: is \: the \: initial \: volume \: occupied \: by \: the \: gas$

$\rm \: P_2  \: is \: the \: final \: pressure \: exerted \: by \: the \: gas$

$\rm \: V_2  \: is \: the \: final \: volume \: occupied \: by \: the \: gas$

$\large\underline\purple{\bold{Solution :- }}$

$\begin{gathered}\begin{gathered}\bf \:Let \: the \: volume \: of \: \begin{cases} &\sf{larger \: container = 3V} \\ &\sf{smaller \: container \: = V} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \:Pressure \: of \: gas \: in\begin{cases} &\sf{large \: container \: is \: 10.3 \: atm} \\ &\sf{small \: container \: is \: 1.1 \: atm.} \end{cases}\end{gathered}\end{gathered}$

Now,

• Initial product of pressure and volume of the system,

$\longrightarrow\tt \: {P_1}\tt{V_1}=\tt{10.3 \times 3V}+ \: 1.1V$

$\longrightarrow\tt \: {P_1}\tt{V_1}=\tt \: 30.9V+ \: 1.1V$

$\longrightarrow\tt \: {P_1}\tt{V_1}=\tt \: 32V$

Now,

• When the two cylinders are connected,

Then,

• The volume of the system becomes V + 3V = 4V.

• Let P be the final pressure of the system.

Then,

• Final product of pressure and volume of the system,

$\longrightarrow\tt{P_2V_2}=4PV$

Now,

$\large \underline{\tt \: \red{ According \: to \: condition}}$

$\rm :\implies\:P_1V_1 = P_2V_2$

$\rm :\implies\:32V = 4PV$

$\rm :\implies\: \boxed{ \red{ \bf \: P = \: 8.0 \: atm.}}$