The pressure and density of a diatomic gas $\left(\mathrm{\gamma } = \frac{7}{5}\right)$ change adiabatically from (P₁ , ρ₁) to (P₂ , ρ₂) . If $\frac{{\mathrm{\rho }}_2}{{\mathrm{\rho }}_1} = 32$ , then $\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}$ should be

• 16

• 32

• 64

• 128

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

• (T − 4) K

• (T + 4) K

• (T − 2.4)K

• (T + 2.4) K

According to first law of thermodynamics

• energy is conserved

• mass is conserved

• heat is constant in isothermal process

• heat neither enters nor leaves system

284.

Ten moles of an ideal gas at constant temperature 500 K is compressed from 50 L to 5 L. Work done in the process is (Given, R = 8.31 J-mol^-1 - K^-1 )

• - 1.2 x 10⁴ J

• - 2.4 x 10⁴ J

• - 4.8 x 10⁴ J
  
  

• - 9.6 x 10⁴ J

An ideal gas is compressed isothermally until its pressure is doubled and then allowed to expand adiabatically to regain its original volume (γ = 1.4 and 2^-1.4 = 0.38). The ratio of the final to initial pressure is

• 0.76 : 1

• 1 : 1

• 0.66 : 1

• 0.86 : 1

The molar specific heats of an ideal gas at constant pressure and volume are denoted by C_P and C_V respectively. If $\mathrm{\gamma } = \frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{V}}}$ and R is the universal gas constant, then C_V is equal to

• $\frac{1 + \mathrm{\gamma }}{1 - \mathrm{\gamma }}$

• $\frac{\mathrm{R}}{\left(\mathrm{\gamma } - 1\right)}$

• $\frac{\left(\mathrm{\gamma } - 1\right)}{\mathrm{R}}$

• γR

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of $\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{v}}}$ for the gas is

• $\frac{4}{3}$

• 2

• $\frac{5}{3}$

• $\frac{3}{2}$

A scientist proposes a new temperature scale in which the ice point is 25 X (X is the new unit of temperature) and the steam point is 305 X. The specific heat capacity of water in this new scale is (in J kg^-1X^-1)

• 4.2 × 10³

• 3.0 × 10³

• 1.2 × 10³

• 1.5 × 10³

One mole of a van der Waals' gas obeying the equation

       $\left(\mathrm{p} + \frac{\mathrm{a}}{{\mathrm{V}}^2}\right) \left(\mathrm{V} - \mathrm{b}\right) =\hspace{0.17em}\mathrm{RT}$

undergoes the quasi-static cyclic process which is shown in the p-V diagram. The net heat absorbed by the gas in this process is

• $\frac{1}{2} \left({\mathrm{p}}_1 - {\mathrm{p}}_2\right) \left({\mathrm{V}}_1 -\hspace{0.17em}{\mathrm{V}}_2\right)$

• $\frac{1}{2} \left({\mathrm{p}}_1 + {\mathrm{p}}_2\right) ({\mathrm{V}}_1 - {\mathrm{V}}_2)$

• $\frac{1}{2} \left({\mathrm{p}}_1 + \frac{\mathrm{a}}{{{\mathrm{V}}_1}^2} - {\mathrm{p}}_2 - \frac{\mathrm{a}}{{{\mathrm{V}}_2}^2}\right) ({\mathrm{V}}_1 - {\mathrm{V}}_2)$

• $\frac{1}{2} \left({\mathrm{p}}_1 + \frac{\mathrm{a}}{{{\mathrm{V}}_1}^2} + {\mathrm{p}}_2 + \frac{\mathrm{a}}{{{\mathrm{V}}_2}^2}\right) ({\mathrm{V}}_1 -{\mathrm{V}}_2)$

A heating element of resistance r is fitted inside an adiabatic cylinder which carries a frictionless piston of mass m and cross-section A as shown in diagram. The cylinder contains one mole of an ideal diatomic gas. The piston current flows through the element such that the temperatures rises with time t as ΔT = αt + $\frac{1}{2}$ βt² ? (α and β are constants), while pressure remains constant. The atmospheric pressure above the piston is P₀. Then

• he rate of increase in internal energy is $\frac{5}{2} \mathrm{R} \left(\mathrm{\alpha } + \mathrm{\beta t}\right)$

• the current flowing in the element is $\sqrt{\frac{5}{2\mathrm{r}} \mathrm{R} \left(\mathrm{\alpha } + \mathrm{\beta t}\right)}$

• the piston moves upwards with constant acceleration

• the piston moves upwards with constant speed