Question

A sample of gas in a box is at pressure $P_0$ and temperature $T_0$. If number of molecules is doubled and total kinetic energy of the gas kept constant then final temperature and pressure will be

• A. $T_0.P_0$
• B. $T_0.2P_0$
• C. $\frac{T_0}{2}.{2P}_0$
• D. $\frac{T_0}{2}.P_0$

Answer 1

Answer: (D)

$\frac{T_0}{2}.P_0$

Total Kinetic Energy = $\frac{3}{2}nRT$

$\therefore$ If kinetic energy is constant;

$\frac{3}{2}{n}_{1}R{T}_1$$=\frac{3}{2}{n}_{2}R{T}_2$

If $n_2=2n_1$

$T_2=\frac{T_1}{2}$ ($\frac{T_0}{2}$)

Again, $\frac{P_1{V}_1}{T_1}=$$\frac{P_2{V}_2}{T_2}$

$P_2=P_1$ $\left(P_0\right)$

$\frac{T_0}{2},$$\left(P_0\right)$ are the final temperature and pressure.