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

We consider a thermodynamic system. If ΔU represents the increase in its internal energy and W the work done by the system, which of the following statements is true ?

1. ΔU = –W in an adiabatic process

2. ΔU = W in an isothermal process

3. ΔU = –W in an isothermal process

4. ΔU = W in an adiabatic process

The volume of a gas is reduced adiabatically to $\frac{1}{4}$ of its volume at 27°C, if the value of γ = 1.4, then the new temperature will be - 

(1) 350 × 4^0.4 K

(2) 300 × 4^0.4 K

(3) 150 × 4^0.4 K

(4) None of these

For an adiabatic expansion of a perfect gas, the value of $\frac{\Delta P}{P}$ is equal to 

(1) $-\sqrt{\gamma }\frac{\Delta V}{V}$

(2) $-\frac{\Delta V}{V}$

(3) $-\gamma \frac{\Delta V}{V}$

(4) $-{\gamma }^2\frac{\Delta V}{V}$

One mole of a perfect gas in a cylinder fitted with a piston has a pressure P, volume V and temperature 273 K. If the temperature is increased by 1 K keeping pressure constant, the increase in volume is

(1) $\frac{2V}{273}$

(2) $\frac{V}{91}$

(3) $\frac{V}{273}$

(4) V

A unit mass of a liquid with volume V₁ is completely changed into a gas of volume V₂ at a constant external pressure P and temperature T. If the latent heat of evaporation for the given mass is L, then the increase in the internal energy of the system is: 
1.  Zero
2. $P(V_2-V_1)$
3. $L-P(V_2-V_1)$
4.  L