The latent heat of vaporisation of water is \(2240~\text{J/gm}\). If the work done in the process of expansion of \(1~\text{g}\) is \(168~\text{J}\), then the increase in internal energy is:
1. \(2408~\text{J}\)
2. \(2240~\text{J}\)
3. \(2072~\text{J}\)
4. \(1904~\text{J}\)

If $\gamma$ denotes the ratio of two specific heats of a gas, the ratio of slopes of adiabatic and isothermal PV curves at their point of intersection is 

(1) $\frac{1}{\gamma }$

(2) $\gamma$

(3) $\gamma -1$

(4) $\gamma +1$

Air in a cylinder is suddenly compressed by a piston, which is then maintained at the same position. With the passage of time 

(1) The pressure decreases

(2) The pressure increases

(3) The pressure remains the same

(4) The pressure may increase or decrease depending upon the nature of the gas

A polyatomic gas \(\left(\gamma = \frac{4}{3}\right)\) is compressed to \(\frac{1}{8}\) of its volume adiabatically. If its initial pressure is \(P_0,\) its new pressure will be:

For adiabatic processes $\left(\gamma =\frac{C_p}{C_v}\right)$ 

(1) $P^{\gamma }V$ = constant

(2) $T^{\gamma }V$ = constant

(3) $T{V}^{\gamma -1}$ = constant

(4) $T{V}^{\gamma }$ = constant

One mole of helium is adiabatically expanded from its initial state $(P_i,V_i,T_i)$ to its final state $(P_f,V_f,T_f)$. The decrease in the internal energy associated with this expansion is equal to

(1) $C_V(T_i-T_f)$

(2) $C_P(T_i-T_f)$

(3) $\frac{1}{2}(C_P+C_V)(Ti-T_f)$

(4) $(C_P-C_V)(T_i-T_f)$

A diatomic gas initially at 18°C is compressed adiabatically to one-eighth of its original volume. The temperature after compression will be 

(1) 10°C

(2) 887°C

(3) 668 K

(4) 144°C

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio C_p/C_v for the gas is 

(1) $\frac{3}{2}$

(2) $\frac{4}{3}$

(3) 2

(4) $\frac{5}{3}$

One mole of an ideal gas at an initial temperature of T K does 6 R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be -

(1) (T + 2.4)K

(2) (T – 2.4)K

(3) (T + 4)K

(4) (T – 4)K