Question

A vessel contains a gas of density $\rho$. The r.m.s velocity of the molecules of the gas is C. If the vessel is moved with a velocity V, then the pressure exerted by the gas is given by
$A.\,P=\dfrac{1}{3}\rho {{[C+V]}^{2}}$
$B.\,P=\dfrac{1}{3}\rho {{C}^{2}}$
$C.\,P=\dfrac{1}{3}\rho {{[C-V]}^{2}}$
$D.\,P=\dfrac{1}{3}\rho [{{C}^{2}}-{{V}^{2}}]$

Answer 1

The formula for calculating the rms voltage and ideal gas law should be used to solve this problem. From the ideal gas obtain the expression for RT and from the mass formula, obtain the expression in terms of the density. Finally, substitute the given values and the obtained values in the rms voltage formula, to obtain the expression in terms of the pressure exerted by the gas.

Formula used:
${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$

Complete step-by-step answer:
The formulae used are as follows.
The ideal gas is given as follows.
$PV=nRT$
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature.
For the specific volume, V, the ideal gas law can be expressed as,
$PV=RT$…… (1)
The rms voltage formula is given as follows.
${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$…… (2)
Where M is the mass, R is the gas constant and T is the temperature.
Substitute the equation (1) in the equation (2)
${{v}_{rms}}=\sqrt{\dfrac{3PV}{M}}$
Express the mass of the gas in terms of the density and volume. So, we have,
$\text{Mass}=\text{Volume}\times \text{Density}$
Substitute the given values in the above equation.
$\text{M}=\text{V}\times \rho$…… (3)
Substitute the equation (3) in the above equation.
${{v}_{rms}}=\sqrt{\dfrac{3PV}{\text{V}\times \rho }}$
Cancel out the common terms
${{v}_{rms}}=\sqrt{\dfrac{3P}{\rho }}$
Substitute the value of the rms voltage given in the above equation and express the equation in terms of the pressure exerted by the gas.
$C=\sqrt{\dfrac{3P}{\rho }}$
Square on both sides of the equation.

& {{C}^{2}}=\dfrac{3P}{\rho } \\\ & \Rightarrow P=\dfrac{1}{3}\rho {{C}^{2}} \\\ \end{aligned}$$ As the expression for the pressure exerted by the gas obtained is equal to $$\dfrac{1}{3}\rho {{C}^{2}}$$, thus, the option (B) is correct. **So, the correct answer is “Option *B)”.** **Note:** This is a direct question, as we have substituted the values in the formulae to obtain the expression of the pressure exerted by the gas. Even they can ask for the expression of the volume of the gas by giving the mass of the gas instead, giving the mass of the gas in terms of density.