Question

If rms velocity of a gass is $V_{rms}=1840m/s$ and its density $\rho =8.99\times {10}^{-2}kg/m^3$, the pressure of the gas will be :-

• A. $1.01N/m^2$
• B. $1.01\times {10}^{3}N/m^2$
• C. $1.01\times {10}^{5}N/m^2$
• D. $1.01\times {10}^{7}N/m^2$

Answer 1

The correct option is C $1.01\times {10}^{5}N/m^2$

We know that $V_{rms}=\sqrt[2]{2\frac{3P}{\rho }}$

then the pressure will be given by $P=\frac{1}{3}\rho V_{rms}^2$

So putting corresponding values , we get $P=1.01\times {10}^{5}N/m^2$

Option C is correct.