Answer: Option B. -> 2nd Law only
:
B
This would violate the $2^{nd}$ Law only. The entropy of the hot reservoir decreases by $\Delta S=\frac{Q}{T}$. The entropy of the engine did not change (since it returns to its initial state). The entropy of the rest of the universe did not change, since doing pure work does not change entropy.

Answer: Option C. -> −4.68 × 105 J
:
C
The change in internal energy $\Delta E_{int}=n{C}_V\Delta T$. We need to obtain $C_V$ from the given information. Now,
$\gamma =\left[\frac{C_P}{C_V}\right]$, and, $C_P-C_V=R$
$\Rightarrow \gamma =\left[\frac{C_V+R}{C_V}\right]=[1+\frac{R}{C_V}]$
$\Rightarrow \frac{R}{C_V}=\gamma -1$
$\Rightarrow C_V=\left[\frac{R}{\gamma -1}\right]$
Therefore,
$\Delta E_{int}=n{C}_V\Delta T$
$=2500\times \left(\frac{8.314}{1.4-1}\right)\times (26-35)J$
$=-4.68\times {10}^{5}J$.
It's a negative change, meaning the internal energy has decreased with cooling, which is consistent with theory. In fact, if an AC had to do this cooling, its job would be take $4.68\times {10}^{5}J$from the internal energy in the air and dump it outside. This will be dealt with in greater detail when we study the second law of thermodynamics.

Answer: Option A. -> 1 : 1
:
A
Since both the gases are diatomic, each has two degree of freedom associated with rotational motion. According to the law of equipartition of energy, the rotational kinetic energy per degree of freedom is $\left(\frac{1}{2}\right)KT$. Since the temperatures of the two gases are equal, their rotational kinetic energies will be equal. Hence the correct choice is (a)

Answer: Option C. -> 1−(1088)0.4
:
C
The efficiency of an Otto engine is given by the formula
$e=1-\frac{1}{r^{\gamma -1}}r=$ compression ratio
Substituting for known values we get
$e=1-\frac{1}{(\frac{88}{10}{)}^{1.4-1}}$
$=1-\frac{1}{(\frac{88}{10}{)}^{.4}}=1-{\left(\frac{10}{88}\right)}^{0.4}$

Answer: Option B. -> 831 J
:
B
The bubble displaces some water when it expands, thus doing some work W, which according to the first law of thermodynamics equals -
$W=Q-\Delta E_{int}.$
Since the expansion process occurs at a constant pressure we have to use the molar specific heat at constant pressure $C_P$, to find the heat gained $\Delta Q$. On the other hand, $\Delta E_{int}$ always depends on the molar specific heat at constant volume $C_V$.
In kinetic theory, $C_P$and $C_V$are related as -
$C_P-C_V=R$
$\Rightarrow C_P=C_V+R$.
Therefore,
$W=Q-\Delta E_{int}$
$=n{C}_P\Delta T-n{C}_V\Delta T$
$=n(C_P-C_V)\Delta T$
$=nR\Delta T$
$=(5\times 8.314\times 20)J$
= 831.4 J.
Note that the work done in a constant-pressure, volume-expansion process is just $W=P\Delta V$, which for an ideal gas, is just $nR\Delta T$, thus in agreement with the result we obtained from the first law.

Answer: Option A. -> 1.72 K
:
A
The oxygen molecules can undergo translational motion in three unique dissections (along x, y and z axes), and rotate about two different axes (as encountered in the previous problem). It, therefore, has five degrees of freedom in total.
Since, according to the equipartition theorem,
$C_V=\left[\frac{numberofdegreesoffreedom}{2}\right]\times R$,
and we already know,
$C_P-C_V=R$,
We can write $C_P$ as -
$C_P=C_V+R$
= $\frac{5}{2}R+R$
= $\frac{7}{2}R$.
$\therefore Q=n{C}_P\Delta T$
$\Rightarrow \Delta T=\frac{Q}{n{C}_P}$
$\Rightarrow \Delta T=\left[\frac{500}{10\times \frac{7}{2}\times 8.314}\right]K$
$\Rightarrow \Delta T=1.72K$.

Answer: Option D. -> 747 J ; 1.29
:
D
Listing the given information,
(i) Number of moles, n = 1.75mol,
(ii) Heat gained, Q = 970J,
(iii) Work done by the gas, W = 223J,
(iv) Rise in temperature,$\Delta T={25}^{\circ }C-{10}^{\circ }C={15}^{\circ }C$.
Therefore, from the first law, gain of internal energy -
$\Delta E_{int}=Q-W$
$\Rightarrow \Delta E_{int}=970J-223J=747J.$
We know, from the kinetic theory, $\Delta E_{int}=n{C}_V\Delta T$
$\Rightarrow C_V=\frac{\Delta E_{int}}{n\Delta T}$
= $\frac{747J}{1.75mol\times {15}^{\circ }C}$
= 28.46 J/mol.K.
Now,
$\gamma =\frac{C_P}{C_V}$
=$\frac{C_V+R}{C_V}$
= $1+\frac{R}{C_V}$
= $1+\frac{8.314}{28.46}=1.29$.

Answer: Option B. -> 2nd law of Thermodynamics
:
B
Efficiency equal to one is allowed by the $1^{st}$ law, as it would mean heat input = heat output. But, it violates the $2^{nd}$ law, wherein work output cannot be achieved by only heat input from a source at single temperature, some heat is always lost to a sink at another temperature, such that the change in entropy of the universe is equal to or greater than zero.

Answer: Option C. -> Both (a) and (b)
:
C
Efficiency greater than 1 means that the work output of the engine is greater than the heat input! This violates the conservation of energy, or the 1${}^{st}$ law of thermodynamics. Also the 2${}^{nd}$ law prohibits the efficiency to be equal to 1, hence it is also being violated.

Answer: Option D. -> No, the efficiency of Carnot's engine is the theoretical upper limit of efficiency, he can at best match it.
:
D
The efficiency of a Carnot's engine is the highest possible theoretical efficiency of any engine, any other engine can only match it, not improve upon it. Any engine with a better efficiency can be proved to violate the 2${}^{nd}$ law of thermodynamics.