
A gas at a pressure P is contained in a vessel of the mass of all the molecules are halved and their velocities doubled the resulting pressure P will be?
Answer
565.8k+ views
Hint: We know that a gas's pressure is directly proportional to the root of the product of its mass and square of root mean square velocity. We will write the equations of initial and final pressure to establish a relationship between them.
Complete step by step answer:
Assume:
The initial mass of all the molecules of a gas is m.
The final mass of all the molecules of a gas is \[{m_2}\].
The initial velocity of molecules is v.
The final velocity of molecules is \[{v_2}\].
The final pressure of the gas is \[{P_2}\].
Let us write the expression for pressure P of the given gas.
\[P = \dfrac{1}{3}\dfrac{{mN}}{V}v_{rms}^2\]
Here, m is mass, N is the number of moles, V is volume, and \[{v_{rms}}\] is the gas's root mean square velocity.
The number of moles and volume of the gas is kept constant so that we can write:
\[P \propto mv_{rms}^2\]
Root mean square velocity of a gas is equal to all the molecules; therefore, we can substitute v for \[{v_{rms}}\] in the above expression.
\[P \propto m{v^2}\]……(1)
We can write the expression for final pressure of the gas as below:
\[{P_2} \propto m{}_2v_2^2\]……(2)
We are given that mass gas is halved, and its velocity is doubled so that we can write:
\[{m_2} = \dfrac{m}{2}\]
And,
\[{v_2} = 2v\]
On substituting \[\dfrac{m}{2}\] for m and 2v for \[{v_2}\] in equation (2), we get:
\[\begin{array}{l}
{P_2} \propto \left( {\dfrac{m}{2}} \right){\left( {2v} \right)^2}\\
\Rightarrow {P_2} \propto 2m{v^2}
\end{array}\]……(3)
On dividing equation (2) and equation (3), we get:
\[\begin{array}{l}
\dfrac{{{P_2}}}{P} = \dfrac{{2m{v^2}}}{{m{v^2}}}\\
\Rightarrow \dfrac{{{P_2}}}{P} = 2
\end{array}\]
Therefore, we can say that the gas's pressure will be doubled if all the molecules' mass is halved, and their velocities are doubled.
Note:Physically, the pressure is created due to the striking of molecules over a surface. Due to striking, there is a change in the momentum, and from Newton's second law, this change in momentum will generate a force that will create pressure.
Complete step by step answer:
Assume:
The initial mass of all the molecules of a gas is m.
The final mass of all the molecules of a gas is \[{m_2}\].
The initial velocity of molecules is v.
The final velocity of molecules is \[{v_2}\].
The final pressure of the gas is \[{P_2}\].
Let us write the expression for pressure P of the given gas.
\[P = \dfrac{1}{3}\dfrac{{mN}}{V}v_{rms}^2\]
Here, m is mass, N is the number of moles, V is volume, and \[{v_{rms}}\] is the gas's root mean square velocity.
The number of moles and volume of the gas is kept constant so that we can write:
\[P \propto mv_{rms}^2\]
Root mean square velocity of a gas is equal to all the molecules; therefore, we can substitute v for \[{v_{rms}}\] in the above expression.
\[P \propto m{v^2}\]……(1)
We can write the expression for final pressure of the gas as below:
\[{P_2} \propto m{}_2v_2^2\]……(2)
We are given that mass gas is halved, and its velocity is doubled so that we can write:
\[{m_2} = \dfrac{m}{2}\]
And,
\[{v_2} = 2v\]
On substituting \[\dfrac{m}{2}\] for m and 2v for \[{v_2}\] in equation (2), we get:
\[\begin{array}{l}
{P_2} \propto \left( {\dfrac{m}{2}} \right){\left( {2v} \right)^2}\\
\Rightarrow {P_2} \propto 2m{v^2}
\end{array}\]……(3)
On dividing equation (2) and equation (3), we get:
\[\begin{array}{l}
\dfrac{{{P_2}}}{P} = \dfrac{{2m{v^2}}}{{m{v^2}}}\\
\Rightarrow \dfrac{{{P_2}}}{P} = 2
\end{array}\]
Therefore, we can say that the gas's pressure will be doubled if all the molecules' mass is halved, and their velocities are doubled.
Note:Physically, the pressure is created due to the striking of molecules over a surface. Due to striking, there is a change in the momentum, and from Newton's second law, this change in momentum will generate a force that will create pressure.
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