The temperature of an open room of volume \(30~\text{m}^3\) increases from \(17^\circ \text{C}\) to \(27^\circ \text{C}\) due to the sunshine. The atmospheric pressure in the room remains \(1\times 10^{5}~\text{Pa}\). In \(n_i\) and \(n_f\) are the number of molecules in the room before and after heating, the \(n_f\text-n_i \) will be:
1. \( -1.61 \times 10^{23} \)
2. \( 1.38 \times 10^{23} \)
3. \( 2.5 \times 10^{25} \)
4. \( -2.5 \times 10^{25}\)

One mole of an ideal gas undergoes a process in which pressure and volume are related by the equation:
        \(P=P_0\left[1-\dfrac{1}{2}\left(\dfrac{V_0}{V}\right)^2\right] \)
where \(P_0\)​ and \(V_0\)​ are constants. If the volume of the gas increases from \(V=V_0\)​ to \(V=2V_0,\) what is the resulting change in temperature?
1. \( \frac{3}{4} \frac{P_o V_o}{R} \)
2. \(\frac{1}{2} \frac{P_o V_o}{R} \)
3. \(\frac{5}{4} \frac{P_o V_o}{R} \)
4. \(\frac{1}{4} \frac{P_o V_o}{R}\)

Number of molecules in a volume of \(4~\text{cm}^3\) of a perfect monoatomic gas at some temperature \(T\) and at a pressure of \(2~\text{cm}\) of mercury is close to ? (Given, mean kinetic energy of a molecule (at \(T\)) is \(4 \times 10^{-14}\)erg, \(g=980\) cm/s² , density of mercury = \(13.6~ \text{g/cm}^3\))
1. \( 5.8 \times 10^{18} \)
2. \( 5.8 \times 10^{16} \)
3. \( 4.0 \times 10^{18} \)
4. \( 4.0 \times 10^{16}\)

Initially, a gas of diatomic molecules is contained in a cylinder of volume \(V_1\) at a pressure \(P_1\) and temperature \(250~\text{K}.\) Assume that \(25\%\) of the molecules get dissociated, causing a change in the number of moles. The pressure of the resulting gas at temperature \(2000~\text{K},\) when contained in a volume \(2V_1\) is given by \(P_2.\) The ratio \(P_2/P_1\) is:
1. \(2\)
2. \(3\)
3. \(5\)
4. \(9\)

On the basis of kinetic theory of gases, the gas exerts pressure because its molecules: