For a diatomic gas,the change in internal energy at constant pressure and the change in internal energy for a unit temperature change are $U_1$ and $U_2$ respectively. Then $U_1 : U_2$ is equal to:

One mole of a monoatomic ideal gas undergoes a quasistatic process, which is depicted by a straight line joining points $(V_{0}, T_{0})$ and $(2 V_{0}, 3 T_{0})$ in a $V-T$ diagram. What is the value of the heat capacity of the gas at the point $(V_{0}, T_{0})$?

Two containers of equal volume contain the same gas at pressures $P_1$ and $P_2$ and absolute temperatures $T_1$ and $T_2$ respectively. On joining the vessels,the gas reaches a common pressure $P$ and common temperature $T$. The ratio $P/T$ is equal to

$A$ horizontal uniform glass tube of $100 \, cm$ length, sealed at both ends, contains a $10 \, cm$ mercury column in the middle. The temperature and pressure of the air on either side of the mercury column are $81^{\circ} C$ and $76 \, cm$ of mercury, respectively. If the air column at one end is kept at $0^{\circ} C$ and the other end at $273^{\circ} C$, the pressure of the air which is at $0^{\circ} C$ is (in $cm$ of $Hg$):

An ideal gas at pressure $P$ and temperature $T$ is expanding such that $PT^3 = \text{constant}$. The coefficient of volume expansion of the gas is . . . . . . .

$CO_2$ $(O-C-O)$ is a triatomic gas. The mean kinetic energy of one gram of the gas will be (where $N$ is Avogadro's number,$k$ is Boltzmann's constant,and the molecular weight of $CO_2 = 44$).