Question

In a container of capacity 1 litre there are ${10}^{23}$ molecules each of mass ${10}^{-22}\text{g}$. If the root mean square speed is ${10}^5\text{cm/s}$ then, calculate the pressure of the gas.

• A. $\frac{1}{3}\times {10}^7\text{Pa}$
• B. $1\times {10}^7\text{Pa}$
• C. $1\text{atm}$
• D. $1\text{torr}$

Answer 1

The correct option is A $\frac{1}{3}\times {10}^7\text{Pa}$

We know the root mean square speed is given as:
$u_{rms}=\sqrt{\frac{3RT}{M}}$
or using $PV=nRT$
$u_{rms}=\sqrt{\frac{3PV}{nM}}$
$u_{rms}=\sqrt{\frac{3PV}{w}}$ (using $n=w/M$)
now,
$PV=\frac{1}{3}w{U}_{\text{rms}}^2$
we have $N={10}^{23}$ molecules so total mass becomes $m\times N$
$PV=\frac{1}{3}mN{U}_{\text{rms}}^2$

$P=?$; $V=1L={10}^{-3}{\text{m}}^3$
$m={10}^{-25}\text{kg}$;
$N={10}^{23}mo{l}^{-1}$
$u_{rms}={10}^5\text{cm/s}={10}^3\text{m/s}$
$u_{rms}^2={10}^6{\text{m}}^2/{\text{s}}^2$

$P\times {10}^{-3}=\frac{1}{3}\times {10}^{-25}\times {10}^{23}\times {10}^6$
$P=\frac{1}{3}\times {10}^{-2}\times {10}^6\times {10}^3$ $P=\frac{1}{3}\times {10}^7\text{Pascals}$