Question 828: Thermodynamics & KTG

Question

A gas at pressure $$p_0$$ is contained in a vessel. If the masses of all the molecules are halved and their velocities are doubled, then the resulting pressure would be equal to

Accepted Answer

Pressure $$p$$ of the gas is given by,
$$p=\frac{1}{3} \frac{m N}{V} v_{\mathrm{rms}}^2$$
Where, $$m$$ is mass, $$N$$ is the number of moles, $$V$$ is the volume and $$v_{\mathrm{rms}}$$ is the root mean square velocity of the particles of the gas.
$$	herefore \quad p_1=p_0=\frac{1}{3} \frac{m_1 N}{V_1}\left(v_{\mathrm{rms}}^2ight)_1$$ ..... (i)
If $$m_2=\frac{m_1}{2}$$ and $$\left(v_{\mathrm{rms}}ight)_2=2\left(v_{\mathrm{rms}}ight)_1$$
For simplicity write $$v_2=2 v_1$$
Hence, $$p_2=\frac{1}{3} \frac{m_2 N}{V} v_2^2$$ ..... (ii)
Divide Eq. (i) by Eq. (ii)
$$\begin{aligned}
& \frac{p_0}{p_2}=\frac{\frac{1}{3} \frac{m_1 N}{V} v_1^2}{\frac{1}{3} \frac{m_2 N}{V} v_2^2} \
& \Rightarrow \quad \frac{p_0}{p_2}=\frac{m_1 v_1^2}{\frac{m_1}{2}\left(2 v_1ight)^2} \Rightarrow \frac{p_0}{p_2}=\frac{1}{2} \
& \Rightarrow \quad p_2=2 p_0 \
\end{aligned}$$
$$	herefore$$ Resulting pressure is $$2 p_0$$

Answer

$$4 p_0$$

Answer

$$2 p_0$$

Answer

$$p_0$$

Answer

$$p_0 / 2$$

Thermodynamics & KTG

A gas at pressure $p_0$ is contained in a vessel. If the masses of all the molecules are halved and their velocities are doubled, then the resulting pressure would be equal to

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