Nitrogen gas is at a certain temperature \(300^\circ \text{C}.\) At what temperature (in Kelvin) will the root mean square (rms) speed of a hydrogen molecule be equal to the rms speed of a nitrogen molecule?
(given: molar mass of nitrogen molecule is \(28~\text g/ \text{mol}\) and molar mass of hydrogen molecule is \(2~\text g/ \text{mol}\))
1. \(21\) K
2. \(41\) K
3. \(52\) K
4. \(76\) K

Match the \(C_p/C_V\)  ratio for ideal gases with different types of molecules:

Given below are two statements :

In the light of the above statements, choose the correct answer from the options given below :
 

Two moles of an ideal monoatomic gas occupy a volume \(V\) at \(27^\circ \text{C}\). The gas expands adiabatically to a volume \(2V\). Calculate the final temperature of the gas and the change in its internal energy:
1. \(189~\text{K}, ~2.7~\text{kJ}~\)
2. \(195~\text{K}, ~-2.7~\text{kJ}~\)
3. \(189~\text{K}, ~-2.7~\text{kJ}~\)
4. \(195~\text{K}, ~2.7~\text{kJ}~\)

If the ratio of the number density per cm³ of the two gases is \(5:3\) and the ratio of the diameters of the molecules of the two gases is \(4:5,\) then, the ratio of the mean free path of molecules of two gases is:

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

The root mean square speed of molecules of a given mass of a gas at \(27^\circ \text{C}\) and \(1~\text{atm}\) is \(200~\text{m/s}\) The root mean square speed of molecules of the gas at \(127^\circ \text{C}\) and \(2~\text{atm}\) will be:

Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom. The gas is maintained at a temperature of \(T.\) The total internal energy, \(U\) of a mole of this gas, and the value of \(\gamma~\left(=\dfrac{C_P}{C_V}\right )\) are, respectively:
1. \( U=5 R T \text { and } \gamma=\dfrac{7}{5} \)

2. \( U=\dfrac{5}{2} R T \text { and } \gamma=\dfrac{6}{5} \)

3. \(U=5 R T \text { and } \gamma=\dfrac{6}{5} \)

4. \( U=\dfrac{5}{2} R T \text { and } \gamma=\dfrac{7}{5}\)

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. \(9\)

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. \( -2.5 \times 10^{25}\)