Physics M.C.Q (1 Marks)

1. heat is completely converted to mechanical energy in such a process, which is not possible.
2. curves representing two adiabatic processes don’t intersect.

Explanation:

1. The given process is a cyclic process, i.e. it returns to the original state 1. And change in internal energy in a cyclic process is always zero as for cyclic process$\text{U}_\text{f}=\text{U}_\text{i}\ \text{So},\Delta\text{U}=\text{U}_\text{f}-\text{U}_\text{i}=0$ Hence, total heat is completely converted to mechanical energy. Such a process is not possible by second law of thermodynamics.
2. Here, two curves are intersecting, when the gas expands adiabatically from 2 to 3. It is not possible to return to the same state without being heat supplied, hence the process 3 to 1 cannot be adiabatic. So, we conclude that such a process does not exist because curves representing two adiabatic processes do not intersect.

1. $2^{\gamma-1}$

Explanation:

According to the P-V diagram shown for the container A (which is going through isothermal process) and for container B (which is going through adiabatic process).

Both the process involves compression of the gas.

(I) Isothermal compression (gas A) (during 1 → 2)

$\text{P}_{1}\text{V}_{1}=\text{P}_{2}\text{V}_{2}$

$\Rightarrow\text{P}_{0}(2\text{V}_{0})^\gamma=\text{P}_{2}(\text{V}_{0})^\gamma$

$\Rightarrow\text{P}_{0}(2\text{V}_{0})=\text{P}_{2}(\text{V}_{0})$

(ii) Adiabatic compression, (gas B) (during 1 → 2)

$\text{P}_{1}\text{V}_{1} ^\gamma=\text{P}_{2}\text{V}_{2}^\gamma$

$\text{P}_{0}(2\text{V}_{0})^\gamma=\text{P}_{2}(\text{V}_{0})^\gamma$

$\text{P}_{2}=\Big(\frac{2\text{V}_{0}}{\text{V}_{0}}\Big)^\gamma\text{P}_{0}=(2)^\gamma\text{P}_{0}$

Hence $\frac{(\text{P}_{2})_\text{B}}{(\text{P}_{2})_\text{A}}=$ Ratio of final pressure $=\frac{(2)^\gamma\text{P}_{0}}{2\text{P}_{0}}=2^{\gamma-1}$

where, $\gamma$ is ratio of specific heat capacities for the gas.

3. Both (a) and (b).

   Expnation:

   S depends on the mass of the substance and it: temperature.

   Heat capacity, $\text{S} = \frac{\text{Heat consumed by given mass}}{\text{Temperature raised}}$

    If given mass is increased, then S increases.

4. $+4\text{P}_0\text{V}_0$

Explanation:

The direction of arrows is anticlockwise so work done is negative equal to the area of loop $=-(3\text{V}_0-\text{V}_0)(2\text{P}_0-\text{P}_0)=-2\text{P}_0\text{V}_0$ verifies the option (b). New work implies external work is done on the system.

4. PV

Explanation:

Work done = Area of $\Delta\text{ABC}$

$=\frac12\text{BC}\times\text{AB}=\frac12(3\text{V}-\text{V})(\text{2P}-\text{P})$

$\text{W}=\text{PV}$

4. 250K.

Explanation:

Here, T₂ = 300K, $\eta_1=40\%$

$\eta_1=1-\frac{\text{T}_2}{\text{T}_1}$

$\frac{40}{100}=1-\frac{300}{\text{T}_1}$

$\therefore\text{T}_1=500\text{K}$

$\eta_2=40\%+\frac{50}{100}\times40\%=60\%$

$\text{T}_2=300\text{K},\text{T}_1=?$

As, $\eta_2=1-\frac{\text{T}'_2}{\text{T}'_1}$

$\therefore\frac{60}{100}=1-\frac{300}{\text{T}'_1}$ or $\text{T}'_1=750\text{K}$

$\text{T}'_1-\text{T}_1=250\text{K}$

3. $\infty$

1. ​​​​​​$\text{Q}_1>\text{Q}_2>0$

3. $\text{Q}_2<\text{Q}_1<0$

Explanation:

From $\text{Q}_{1}=\text{W}+\text{Q}_{2}$ 

$\therefore\text{W}>0$ So $\therefore\text{Q}_{1}-\text{Q}_{2}$ or $\text{Q}_{1}>0$

$\therefore\text{Q}_{1}>\text{Q}_{2}>0$ if both $\text{Q}_{1},\text{Q}_{2}$ position verifies option (a).

Or $\text{Q}_{2}<\text{Q}_{1}<0$ if both $\text{Q}_{1}\text{Q}_{2}$ negative verifies option (c).

1. dU = 0

4. dQ = dW

Explanation:

Key concept: First Law of Thermodynamics.

It is a statement of conservation of energy in thermodynamical process.

According to it heat given to a system $(\Delta\text{Q})$ is equal to the sum of increase in its internal energy (AIT) and the work done (AW) by the system against the surroundings.

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

According to the first law of thermodynamics. $\Delta\text{A}\text{Q}=\Delta\text{U}+\Delta\text{W}$ but

$\Delta\text{U}\propto\Delta\text{T}$

$\Delta\text{U}=0[\text{As } \Delta\text{T}=0]$

$\Delta\text{Q}=\Delta\text{W}$, i.e., heat supplied in an isothermal change is used to do work against external surrounding or if the work is done on the system then equal amount of heat energy will be liberated by the system

4. $\frac32$

Explanation:

For an adiabatic change,

$\text{PT}^{\frac{\gamma}{(1-\gamma)}}=$ constant

$\therefore\frac{\gamma}{1-\gamma}=-3$ or $-3+3\gamma=\gamma$

$2\gamma=3$

$\gamma=\frac32$

3. $\eta=\frac{\text{W}}{\text{Q}_1}$

Explanation:

For heat engine, Q₁ = W + Q₂

⇒ W = Q₁ - Q₂

Q₂ is heat rejected to the environment.

So, $\eta=\frac{\text{Q}_1-\text{Q}_2}{\text{Q}_1}=\frac{\text{W}}{\text{Q}_1}$

3. 0.577

Explanation:

$\eta_1=1-\frac{273+0}{200+273}=\frac{200}{473}$

$\eta_2=1-\frac{-200+273}{0+27 3}=\frac{200}{273}$

$\frac{\eta_1}{\eta_2\text{}}=\frac{200}{473}\times\frac{273 }{200}=\frac{273}{473}=0.577$

3. $\text{Diatomic}\frac72\text{R},\frac52\text{R}$

2. $\text{P}$

2. $\text{T}=\frac{\text{M}_{1}\text{T}_{1}+\text{M}_{2}\text{T}_{2}+\text{M}_{3}\text{T}_{3}}{\text{M}_{1}+\text{M}_{2}+\text{M}_{3}}$

Explanation:

Let the equilibrium tempreature of the system = T

Let $\text{T}_{1}\text{T}_{2}<\text{T}<\text{T}_{3}$

As there is no loss to the surrounding.

Heat lost by $\text{M}_{3}=$ Heat gain by $\text{M}_{1}$+Heat gain by $\text{M}_{2}$

$\text{M}_{3}\text{s}(\text{T}_{3}-\text{T})=\text{M}_{1}\text{s}(\text{T}-\text{T}_{1})+\text{M}_{2}\text{s}(\text{T}-\text{T}_{2})$

$\text{M}_{3}\text{s}\text{T}_{3}-\text{M}_{3}\text{s}\text{T}=\text{M}_{1}\text{s}\text{T}=\text{M}_{1}\text{s}\text{T}-\text{M}_{1}\text{s}\text{T}_{1}+\text{M}_{2}\text{s}\text{T}-\text{M}_{2}\text{s}\text{T}_{2}$

$\text{T}(\text{M}_{3}+\text{M}_{1}+\text{M}_{2})=[\text{M}_{3}\text{T}_{3}+\text{M}_{1}\text{T}_{1}+\text{M}_{2}\text{T}_{2}]$

$\text{T}=\frac{\text{M}_{1}\text{T}_{1}+\text{M}_{2}\text{T}_{2}+\text{M}_{3}\text{T}_{3}}{\text{M}_{1}+\text{M}_{2}+\text{M}_{3}}$

Hence verifies option (b).

4. $\frac72$

Explanation:

Equation of adiabatic change is,

$\therefore\text{P}\propto\text{T}^{\frac{-\gamma}{(1-\gamma)}}$

For diatomic gas, $\gamma=\frac75$

$\text{P}\propto\text{T}^{\frac72}$

4. $\frac72$

3. $\mu\text{RT}\log\frac{\text{V}_2}{\text{V}_1}$

4. $\text{Both (b) and (c)}.$

Explanation:

For isothermal expansion,

$\Delta\text{U}=0$

$\Rightarrow\Delta\text{U}=\Delta\text{W}$

Here, $\Delta\text{W}\rightarrow+\text{ve}$

$\Rightarrow\Delta\text{Q}\rightarrow+\text{ve}$

3. Diatomic, $\frac{7}{2}\text{R},\frac52\text{R}$

Explanation:

From $\gamma=\Big(1+\frac2{\text{n}}\Big)=1.4$

We get n = 5, which is the number of degrees of freedom of a diatomic gas.

$\text{C}_\text{v}=\frac{\text{n}}{2}\text{R}=\frac{5}{2}\text{R}$

$\text{C}_\text{P}=\Big(\frac{\text{n}}{2}+1\Big)\text{R}=\frac{7}{2}\text{R}$

4. All of the above.

Heat flows from body A to body B because temperature of body A is higher. At thermal equilibrium, $\text{T}_\text{A}'=\text{T}_\text{B}'$

1. 26.81%.

Explanation:

T₁ = (100 + 273) = 373K

T₂ = (0 + 273) = 273K

Efficiency $(\eta)=1-\frac{\text{T}_2}{\text{T}_1}$

$=1-\frac{273}{373}=\frac{100}{373}$

$\%\eta=\frac{100}{373}\times100=26.81\%$

2. $\text{P}$

Explanation:

K_i = P = the pressure exerted by the gas.