Correct answer is (D)

For an adiabatic process,

$P{V}^{\gamma }$ = constant …(i)

According to ideal gas equation

$\mathrm{P}\mathrm{V}=\mathrm{n}\mathrm{R}\mathrm{T}\Rightarrow P=\frac{nRT}{V}$

Putting value of P in (i), we get

$\frac{nRT}{V}{V}^{\gamma }$ = constant; $\therefore$ $T{V}^{\gamma -1}$ = constant

Again from the ideal gas equation

$V=\frac{nRT}{P}$

Putting value of V in (i), we get

$P{\left(\frac{nRT}{P}\right)}^{\gamma }$ = constant; $P^{1-\gamma }{T}^{\gamma }$ = constant

Correct answer is (B)

According to question $P\propto T^3$

But as we know for an adiabatic process the pressure $P\propto T^{\frac{\gamma }{\gamma -1}}$

So, $\frac{\gamma }{\gamma -1}=3\Rightarrow \gamma =\frac{3}{2}$

$\therefore$ $\frac{C_p}{C_v}=\frac{3}{2}$

Correct answer is (D)

From the above P-V diagrams, (d) is correct, as from the question, the initial gas goes from volume V to 3V and then volume of the gas get reduced from 3V to V at constant pressure. In case of an isothermal expansion, P-V curve is rectangular hyperbola which is stated by (d).

Correct answer is (D)

T₁ = 273 + 27 = 300K

T₂ = 273 + 927 = 1200K

For adiabatic process,

$P^{1-\gamma }{T}^{\gamma }$ = constant

$\Rightarrow {P_1}^{1-\gamma }{T_1}^{\gamma }={P_2}^{1-\gamma }{T_2}^{\gamma }$

$\Rightarrow {\left(\frac{P_2}{P_1}\right)}^{1-\gamma }={\left(\frac{T_1}{T_2}\right)}^{\gamma }$

$\Rightarrow {\left(\frac{P_1}{T_2}\right)}^{1-\gamma }={\left(\frac{T_2}{T_1}\right)}^{\gamma }$

${\left(\frac{P_1}{P_2}\right)}^{1-1.4}={\left(\frac{1200}{300}\right)}^{1.4}$

${\left(\frac{P_1}{P_2}\right)}^{-0.4}={\left(4\right)}^{1.4}$

${\left(\frac{P_2}{P_1}\right)}^{0.4}=4^{1.4}$

$P_2=P_1{4}^{\left(\frac{1.4}{0.4}\right)}=P_1{4}^{\left(\frac{7}{2}\right)}$

= P₁ (2⁷) = 2 × 128 = 256 atm

Correct answer is (D)

If a process is expansion then work done is positive so answer will be (a).

But in question work done by gas is given –150J so that according to it answer will be (d).

Correct answer is (C)

Ideal gas equation, for an adiabatic process is

$P{V}^{\gamma }=$ constant $\Rightarrow$ $P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

For monoatomic gas $\gamma =\frac{5}{3}$

$\therefore$ $P_1{V}_1^{5/3}=P_2{\left(\frac{V_1}{8}\right)}^{5/3}$

$\Rightarrow$ P₂ = P₁ × (2)⁵ = 32 P₁.

Correct answer is (A)

By first law of thermodynamics,

$\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W$

In adiabatic process, $\mathrm{\Delta }Q=0$

$\therefore$ $\mathrm{\Delta }U=-\mathrm{\Delta }W$

In isothermal process, $\mathrm{\Delta }U=0$

$\therefore$ $\mathrm{\Delta }Q=\mathrm{\Delta }W$

Correct answer is (A)

The only statement which is not true is statement (a). A process in which the pressure remains constant is called isobaric process and not isochoric as in isochoric process the volume remains constant.

Correct answer is (A)

Internal energy depends only on the initial and final states of temperature and not on the path. In a cyclic process, as initial and final states are the same, change in internal energy is zero. Hence E is $\mathrm{\Delta }$U, the change in internal energy.

Correct answer is (C)

For a process to be reversible, it must be quasi-static. For quasi static process, all changes take place infinitely slowly. Isothermal process occur very slowly so it is quasi-static and hence it is reversible.

Correct answer is (D)

$T_1=T,W=6Rjoules,\gamma =\frac{5}{3}$

$W=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=\frac{nR{T}_1-nR{T}_2}{\gamma -1}$

$=\frac{nR\left(T_1-T_2\right)}{\gamma -1}$

$n=1,T_1=T\Rightarrow \frac{R\left(T-T_2\right)}{5/3-1}=6R$

$\Rightarrow T_2=\left(T-4\right)K$