First Law of Thermodynamics Questions in English

(A) According to the first law of thermodynamics: $\Delta Q = \Delta U + \Delta W$.
Here,$\Delta Q = 100\,J$ (heat supplied to the system).
The work done by the gas is $\Delta W = P \Delta V = P(V_f - V_i)$.
Given $P = 50\,N/m^2$,$V_i = 10\,m^3$,and $V_f = 4\,m^3$.
$\Delta W = 50 \times (4 - 10) = 50 \times (-6) = -300\,J$.
Substituting these values into the first law equation:
$100 = \Delta U + (-300)$.
$\Delta U = 100 + 300 = 400\,J$.
Therefore,the increase in internal energy is $400\,J$.

(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta U$ is the change in internal energy,$\Delta Q$ is the heat supplied,and $\Delta W$ is the work done by the system.
Given,$\Delta Q = 20 \, cal = 20 \times 4.2 \, J = 84 \, J$.
The work done $\Delta W$ is given as $50 \, J$. Since the process $A \rightarrow B \rightarrow C$ involves an expansion (or is part of a cycle),we use the convention where work done by the system is positive.
Therefore,$\Delta U = \Delta Q - \Delta W = 84 \, J - 50 \, J = 34 \, J$.

(C) According to the first law of thermodynamics,$Q = \Delta U + W$,where $Q$ is the heat added,$\Delta U$ is the change in internal energy,and $W$ is the work done.
Internal energy is a state function,meaning it depends only on the initial and final states of the system.
In a closed cyclic process,the system returns to its initial state after completing the cycle.
Since the initial and final states are identical,the change in internal energy $\Delta U$ (denoted here as $E$) must be zero.
Therefore,$E = 0$.

(B) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
First,convert the heat energy from calories to Joules: $\Delta Q = 54 \, cal \times 4.18 \, J/cal = 225.72 \, J$.
Next,calculate the work done during expansion: $\Delta W = P \Delta V$.
Given $P = 1.013 \times 10^5 \, N/m^2$ and $\Delta V = 167.1 \, cc = 167.1 \times 10^{-6} \, m^3$.
$\Delta W = 1.013 \times 10^5 \times 167.1 \times 10^{-6} \approx 16.93 \, J$.
Now,calculate the change in internal energy: $\Delta U = \Delta Q - \Delta W$.
$\Delta U = 225.72 \, J - 16.93 \, J = 208.79 \, J$.
Rounding to the nearest option,$\Delta U \approx 208.7 \, J$.

(C) According to the first law of thermodynamics: $\Delta Q = \Delta U + \Delta W$.
Here,the gas releases heat,so $\Delta Q = -30 \ J$.
Work is done on the gas,so $\Delta W = -10 \ J$.
Let the initial internal energy be $U_i = 30 \ J$ and the final internal energy be $U_f$.
Then,$\Delta U = U_f - U_i = U_f - 30$.
Substituting these values into the equation: $-30 = (U_f - 30) + (-10)$.
$-30 = U_f - 30 - 10$.
$-30 = U_f - 40$.
$U_f = 40 - 30 = 10 \ J$.

(B) Heat added at constant volume is equal to the change in internal energy of the system. The change in internal energy is given by $\Delta U = n C_{v} \Delta T$.
For $\Delta T = 150 \ K$,the change in internal energy is $\Delta U = Q_{v} = 6300 \ J$.
Therefore,$6300 = n C_{v} (150) \quad \dots(1)$.
We need to find the change in internal energy $\Delta U'$ for $\Delta T' = 300 \ K$.
Thus,$\Delta U' = n C_{v} (300) \quad \dots(2)$.
Dividing equation $(2)$ by equation $(1)$,we get $\frac{\Delta U'}{6300} = \frac{300}{150} = 2$.
Therefore,$\Delta U' = 6300 \times 2 = 12600 \ J$.

(B) According to the first law of thermodynamics,the change in internal energy is given by $\Delta U = Q - W$.
The work done $W$ by the system during the phase change at constant atmospheric pressure $p_0$ is $W = p_0 \Delta V$.
For a unit mass $(m = 1)$,the volume of water is $V_1 = \frac{1}{\rho_1}$ and the volume of steam is $V_2 = \frac{1}{\rho_2}$.
Thus,the work done is $W = p_0 (V_2 - V_1) = p_0 \left( \frac{1}{\rho_2} - \frac{1}{\rho_1} \right)$.
Substituting this into the first law equation: $\Delta U = Q - p_0 \left( \frac{1}{\rho_2} - \frac{1}{\rho_1} \right) = Q + p_0 \left( \frac{1}{\rho_1} - \frac{1}{\rho_2} \right)$.

(B) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by $\Delta U = Q + W$,where $Q$ is the heat added to the system and $W$ is the work done on the system.
Given that the gas releases $30 \ J$ of heat,$Q = -30 \ J$.
Since $18 \ J$ of work is done on the gas,$W = +18 \ J$.
Therefore,the change in internal energy is $\Delta U = -30 \ J + 18 \ J = -12 \ J$.
The change in internal energy is defined as $\Delta U = U_f - U_i$,where $U_f$ is the final internal energy and $U_i$ is the initial internal energy.
Given $U_i = 60 \ J$,we have $U_f - 60 \ J = -12 \ J$.
Solving for $U_f$,we get $U_f = 60 \ J - 12 \ J = 48 \ J$.

(C) Given: Mass $m = 1\,g = 10^{-3}\,kg$,Pressure $P = 10^5\,Pa$,Volume of steam $V_s = 1841\,cm^3 = 1841 \times 10^{-6}\,m^3$,Latent heat $L = 2250\,J/g = 2250 \times 10^3\,J/kg$.
The volume of $1\,g$ of water is $V_w = \frac{m}{\rho} = \frac{1\,g}{1\,g/cm^3} = 1\,cm^3 = 10^{-6}\,m^3$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + W$,where $W = P(V_s - V_w)$.
Heat absorbed $\Delta Q = m \times L = 1\,g \times 2250\,J/g = 2250\,J$.
Work done $W = P(V_s - V_w) = 10^5\,Pa \times (1841 \times 10^{-6} - 1 \times 10^{-6})\,m^3 = 10^5 \times 1840 \times 10^{-6} = 184\,J$.
Change in internal energy $\Delta U = \Delta Q - W = 2250\,J - 184\,J = 2066\,J$.

(D) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta Q$ is the heat supplied,$\Delta U$ is the change in internal energy,and $\Delta W$ is the work done.
For the first expansion:
$\Delta W_1 = 20 \, kJ$
$\Delta Q_1 = 16 \, kJ$
Using $\Delta U = \Delta Q_1 - \Delta W_1 = 16 \, kJ - 20 \, kJ = -4 \, kJ$.
Since internal energy $U$ is a state function,the change in internal energy $\Delta U$ depends only on the initial and final states. Therefore,$\Delta U$ remains the same for the second expansion.
For the second expansion:
$\Delta U = -4 \, kJ$
$\Delta Q_2 = 9 \, kJ$
Using $\Delta W_2 = \Delta Q_2 - \Delta U = 9 \, kJ - (-4 \, kJ) = 13 \, kJ$.
Thus,the work done in the second expansion is $13 \, kJ$.

(D) For an ideal gas,the internal energy depends only on temperature. Since the temperature of the gas remains unchanged,the change in internal energy $\Delta U = 0$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + W$. Since $\Delta U = 0$,we have $\Delta Q = W$.
The heat supplied by the resistance coil in time $t$ is $\Delta Q = i^2 R t$.
The work done by the gas in moving the piston of mass $m$ upward by a distance $x$ is $W = mgx$ (since atmospheric pressure is neglected).
Equating the two,we get $i^2 R t = mgx$.
Differentiating with respect to time $t$,we get $i^2 R = mg \frac{dx}{dt}$.
Since the speed of the piston is $v = \frac{dx}{dt}$,we have $i^2 R = mgv$.
Therefore,the speed of the piston is $v = \frac{i^2 R}{mg}$.

(C) The first law of thermodynamics is given as $\Delta U = \Delta Q + \Delta W$.
In this convention,$\Delta W$ represents the work done $ON$ the gas.
For option $A$: If no heat enters or leaves,$\Delta Q = 0$,so $\Delta U = \Delta W$. Thus,$\Delta U$ is not necessarily zero.
For option $B$: In the equation $\Delta U = \Delta Q + \Delta W$,$\Delta W$ is the work done on the gas,not by the gas.
For option $C$: For an ideal gas,internal energy $\Delta U$ is a function of temperature only,i.e.,$\Delta U = nC_v \Delta T$. If the temperature remains constant,$\Delta T = 0$,which implies $\Delta U = 0$. This statement is correct.
For option $D$: If temperature increases,$\Delta U \neq 0$,so $\Delta Q + \Delta W \neq 0$.

(B) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is a state function and depends only on the initial and final states,not on the path taken.
For the path $iaf$:
$\Delta U = Q_{iaf} - W_{iaf}$
$\Delta U = 50 \, cal - 20 \, cal = 30 \, cal$.
Since the initial state $i$ and final state $f$ are the same for the path $ibf$,the change in internal energy $\Delta U$ remains the same.
For the path $ibf$:
$\Delta U = Q_{ibf} - W_{ibf}$
$30 \, cal = 36 \, cal - W_{ibf}$
$W_{ibf} = 36 \, cal - 30 \, cal = 6 \, cal$.

(B) The total heat supplied $(\Delta Q)$ is used for two purposes: increasing the internal energy $(\Delta U)$ and doing work against external pressure $(\Delta W)$.
Given:
$\Delta Q = 540 \ \text{cal}$
$P = 1.013 \times 10^5 \ \text{Nm}^{-2}$
$V_1 = 1 \ \text{cc} = 1 \times 10^{-6} \ \text{m}^3$
$V_2 = 1671 \ \text{cc} = 1671 \times 10^{-6} \ \text{m}^3$
$J = 4.18 \ \text{J/cal}$
Work done against atmospheric pressure is $\Delta W = P(V_2 - V_1)$.
$\Delta W = 1.013 \times 10^5 \times (1671 - 1) \times 10^{-6} \ \text{J}$
$\Delta W = 1.013 \times 1670 \times 10^{-1} \ \text{J} \approx 169.17 \ \text{J}$.
Converting work into calories: $\Delta W_{\text{cal}} = \frac{169.17}{4.18} \approx 40.47 \ \text{cal}$.
The heat required to overcome molecular attraction is the change in internal energy $\Delta U$.
$\Delta U = \Delta Q - \Delta W_{\text{cal}}$
$\Delta U = 540 - 40.47 = 499.53 \ \text{cal} \approx 500 \ \text{cal}$.

(A) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is a state function and depends only on the initial and final states,not on the path taken.
For the path $iaf$:
$Q_{iaf} = \Delta U + W_{iaf}$
$50 \ cal = \Delta U + 20 \ cal$
$\Delta U = 50 - 20 = 30 \ cal$
Since the initial state $i$ and final state $f$ are the same for path $ibf$,the change in internal energy $\Delta U$ remains the same.
For the path $ibf$:
$Q_{ibf} = \Delta U + W_{ibf}$
$36 \ cal = 30 \ cal + W_{ibf}$
$W_{ibf} = 36 - 30 = 6 \ cal$

(C) According to the first law of thermodynamics,the heat supplied $(dQ)$ is equal to the sum of the change in internal energy $(dU)$ and the work done $(dW)$: $dQ = dU + dW$.
The heat supplied for vaporization is given by $dQ = m \times L$,where $m$ is the mass and $L$ is the latent heat of vaporization.
Given: $m = 1 \, g$,$L = 2240 \, J/g$,and $dW = 168 \, J$.
Calculating $dQ$: $dQ = 1 \, g \times 2240 \, J/g = 2240 \, J$.
Substituting the values into the first law equation: $2240 \, J = dU + 168 \, J$.
Solving for $dU$: $dU = 2240 \, J - 168 \, J = 2072 \, J$.
Therefore,the increase in internal energy is $2072 \, J$.

(D) The volume of water is assumed to be constant,so no work is done by or on the system $(W = 0)$.
According to the first law of thermodynamics:
$Q = \Delta U + W$
Since the process is isochoric $(W = 0)$,the heat supplied is equal to the change in internal energy:
$Q = \Delta U$
The formula for heat is $Q = mc\Delta T$,where:
$m = 200\,g = 0.2\,kg$
$c = 4184\,J/kg\cdot K$
$\Delta T = 60\,^{\circ}C - 40\,^{\circ}C = 20\,K$
Substituting the values:
$\Delta U = 0.2\,kg \times 4184\,J/kg\cdot K \times 20\,K$
$\Delta U = 16736\,J$
$\Delta U = 16.736\,kJ \approx 16.7\,kJ$.

(C) The work done during the phase change at constant pressure is given by $W = P \Delta V$.
Given $P = 1.013 \times 10^5\,Pa$ (atmospheric pressure),$V_{liquid} = 1\,cm^3 = 10^{-6}\,m^3$,and $V_{vapour} = 1671\,cm^3 = 1671 \times 10^{-6}\,m^3$.
$W = 1.013 \times 10^5 \times (1671 - 1) \times 10^{-6} \approx 169\,J$. Using the standard approximation $P = 10^5\,Pa$ as implied by the options: $W = 10^5 \times 1670 \times 10^{-6} = 167\,J$.
The heat supplied is $Q = mL = 1\,g \times 2256\,J/g = 2256\,J$.
From the first law of thermodynamics,$Q = \Delta U + W$,so $\Delta U = Q - W$.
$\Delta U = 2256\,J - 167\,J = 2089\,J$.

(A) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by $\Delta U = Q - W$,where $Q$ is the heat supplied to the system and $W$ is the work done by the system.
Since the initial state $A$ and final state $B$ are the same for both paths $ACB$ and $ADB$,the change in internal energy $\Delta U$ must be the same for both processes.
For path $ACB$:
$Q_{ACB} = 60 \, J$
$W_{ACB} = 30 \, J$
$\Delta U = Q_{ACB} - W_{ACB} = 60 - 30 = 30 \, J$
For path $ADB$:
$W_{ADB} = 10 \, J$
Since $\Delta U$ is the same,$\Delta U = Q_{ADB} - W_{ADB}$
$30 = Q_{ADB} - 10$
$Q_{ADB} = 30 + 10 = 40 \, J$
Thus,the heat flow into the system in path $ADB$ is $40 \, J$.

(A) According to the first law of thermodynamics,$dQ = dU + dW$,where $dQ$ is the heat supplied to the system,$dU$ is the change in internal energy,and $dW$ is the work done by the system.
Given:
Heat released by the gas,$dQ = -20 \ J$ (negative because heat is released).
Work done on the gas,$dW = -8 \ J$ (negative because work is done on the system).
Initial internal energy,$U_i = 30 \ J$.
Using the formula $dQ = (U_f - U_i) + dW$:
$-20 = (U_f - 30) + (-8)$
$-20 = U_f - 30 - 8$
$-20 = U_f - 38$
$U_f = 38 - 20$
$U_f = 18 \ J$.
Therefore,the final internal energy is $18 \ J$.

(B) According to the first law of thermodynamics,the heat supplied to a system is equal to the sum of the change in internal energy and the work done by the system: $\delta Q = \Delta U + W$.
Here,$\delta Q = -50 \text{ calorie}$ and $W = -20 \text{ calorie}$.
Substituting these values into the equation: $-50 = \Delta U + (-20)$.
Solving for $\Delta U$: $\Delta U = -50 + 20 = -30 \text{ calorie}$.
We know that $\Delta U = U_f - U_i$,where $U_f$ is the final internal energy and $U_i$ is the initial internal energy.
Given $U_i = -30 \text{ calorie}$,we have: $-30 = U_f - (-30)$.
$-30 = U_f + 30$.
$U_f = -30 - 30 = -60 \text{ calorie}$.

(B) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Given that the gas releases $20 \, J$ of heat,the heat supplied is $\Delta Q = -20 \, J$.
Since $8 \, J$ of work is done on the gas,the work done by the gas is $\Delta W = -8 \, J$.
Substituting these values into the first law equation:
$-20 \, J = \Delta U + (-8 \, J)$.
$\Delta U = -20 \, J + 8 \, J = -12 \, J$.
We know that $\Delta U = U_f - U_i$,where $U_f$ is the final internal energy and $U_i$ is the initial internal energy.
Given $U_i = 30 \, J$,we have:
$U_f - 30 \, J = -12 \, J$.
$U_f = 30 \, J - 12 \, J = 18 \, J$.
Therefore,the final internal energy is $18 \, J$.

(B) The first law of thermodynamics is given by the equation $\Delta Q = \Delta U + W_{by}$,where $W_{by}$ is the work done $BY$ the system.
Given that $\Delta W$ is the work done $ON$ the system,we have $W_{by} = -\Delta W$.
Substituting this into the first law equation:
$\Delta Q = \Delta U + (-\Delta W)$
$\Delta Q = \Delta U - \Delta W$
Therefore,the correct option is $B$.

(C) According to the first law of thermodynamics, the relationship between heat $(Q)$, internal energy change $(\Delta U)$, and work done $(W)$ is given by:
$Q = \Delta U + W$
Given:
Heat absorbed $(Q) = 2 \, kcal = 2 \times 1000 \times 4.2 \, J = 8400 \, J$
Work done $(W) = 500 \, J$
Rearranging the formula to find the change in internal energy:
$\Delta U = Q - W$
Substituting the values:
$\Delta U = 8400 \, J - 500 \, J$
$\Delta U = 7900 \, J$

(D) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W = \Delta U + P\Delta V$.
If heat is supplied in such a manner that the volume does not change $(\Delta V = 0)$,i.e.,an isochoric process,then the whole of the heat energy supplied to the system will increase the internal energy only.
However,in any other process where work is done,the heat supplied is distributed between the change in internal energy and the work done by the system.
Therefore,the $Assertion$ is incorrect.
The $Reason$ is also incorrect because when a system changes from one thermal equilibrium to another,heat may be absorbed,evolved,or remain zero depending on the process.

(B) According to the first law of thermodynamics,$Q = \Delta U + W$.
Here,$Q$ is the heat supplied,$\Delta U$ is the change in internal energy,and $W$ is the work done by the system.
Given:
Mass $m = 1\;g$
Latent heat $L = 2256\;J/g$
Pressure $P = 1 \times 10^{5}\;Pa$
Initial volume $V_{1} = 1\;cm^{3} = 1 \times 10^{-6}\;m^{3}$
Final volume $V_{2} = 1671\;cm^{3} = 1671 \times 10^{-6}\;m^{3}$
Heat supplied $Q = mL = 1\;g \times 2256\;J/g = 2256\;J$.
Work done $W = P(V_{2} - V_{1}) = 1 \times 10^{5} \times (1671 - 1) \times 10^{-6} = 10^{5} \times 1670 \times 10^{-6} = 167\;J$.
Using $Q = \Delta U + W$,we get $\Delta U = Q - W = 2256 - 167 = 2089\;J$.

(D) Heat is supplied to the system at a rate of $100\;W.$
Heat supplied,$Q = 100\;J/s.$
The system performs work at a rate of $75\;J/s.$
Work done,$W = 75\;J/s.$
From the first law of thermodynamics,we have:
$Q = \Delta U + W$
Where,$\Delta U$ is the rate of change of internal energy.
$\therefore \Delta U = Q - W$
$= 100 - 75$
$= 25\;J/s.$
Therefore,the internal energy of the system increases at a rate of $25\;J/s.$

(N/A) Benjamin Thomson,also known as Count Rumford,conducted a famous experiment in $1798$ while observing the boring of cannons.
$1$. He observed that when a blunt drill was used to bore a cannon,a significant amount of heat was generated,which could boil water.
$2$. He noted that the heat production continued as long as the mechanical work of turning the drill was performed.
$3$. Since the heat did not seem to come from the metal itself (as the chips of metal produced were not significantly different in heat capacity),he concluded that heat is not a fluid (caloric) but is instead a form of energy generated by mechanical work.
$4$. This experiment provided early evidence that mechanical work can be converted into heat,laying the groundwork for the first law of thermodynamics.

(N/A) The internal energy of a system can change through two modes of energy transfer: heat and work.
Let $\Delta Q$ = Heat supplied to the system by the surroundings.
Let $\Delta W$ = Work done by the system on the surroundings.
Let $\Delta U$ = Change in the internal energy of the system.
From the principle of conservation of energy,$\Delta Q = \Delta U + \Delta W$ ... $(1)$ is known as the first law of thermodynamics.
From equation $(1)$,it is stated that the energy $(\Delta Q)$ supplied to the system is used partly to increase the internal energy of the system $(\Delta U)$ and the rest is used to perform work on the surroundings $(\Delta W)$.
An alternative form of this equation is $\Delta Q - \Delta W = \Delta U$ ... $(2)$.
$A$ system may transition from an initial state to a final state in various ways. For example,to change the state of a gas from $(P_{1}, V_{1})$ to $(P_{2}, V_{2})$,one could change the volume from $V_{1}$ to $V_{2}$ at constant pressure,then change the pressure from $P_{1}$ to $P_{2}$ at constant volume.
When a system is taken from one state to another,$\Delta Q$ and $\Delta W$ are different for different paths,meaning $\Delta Q$ and $\Delta W$ are path-dependent. However,the difference $(\Delta Q - \Delta W)$ is the same for every path. Thus,$(\Delta Q - \Delta W)$ is path-independent and depends only on the initial and final states of the system.
Therefore,the first law of thermodynamics is $\Delta Q - \Delta W = \Delta U$.
This law is obeyed in all natural processes. All three terms must have the same units.
Sign convention: If the system absorbs heat,$\Delta Q$ is positive; if heat is released,$\Delta Q$ is negative. When a gas expands,work done by the system $\Delta W$ is positive; when compressed,work done on the system $\Delta W$ is negative.
If $(\Delta Q - \Delta W) > 0$,the internal energy increases; if $(\Delta Q - \Delta W) < 0$,the internal energy decreases.

(N/A) Let us consider $1 \,g$ of water for conversion into steam.
Latent heat of vaporisation of water, $L_{v} = 2256 \,J/g$.
Therefore, the heat absorbed is $\Delta Q = m \times L_{v} = 1 \,g \times 2256 \,J/g = 2256 \,J$.
Volume of $1 \,g$ water at atmospheric pressure is $V_{l} = 1 \,cm^{3} = 1 \times 10^{-6} \,m^{3}$.
Volume of $1 \,g$ steam is $V_{g} = 1671 \,cm^{3} = 1671 \times 10^{-6} \,m^{3}$.
Change in volume, $\Delta V = V_{g} - V_{l} = (1671 - 1) \times 10^{-6} \,m^{3} = 1670 \times 10^{-6} \,m^{3}$.
Work done at constant atmospheric pressure, $\Delta W = P \Delta V = (1.013 \times 10^{5} \,Pa) \times (1670 \times 10^{-6} \,m^{3}) \approx 169.2 \,J$.
From the first law of thermodynamics, $\Delta U = \Delta Q - \Delta W$.
$\Delta U = 2256 \,J - 169.2 \,J = 2086.8 \,J$.
Thus, the change in internal energy is $2086.8 \,J$.

(B) The internal energy $(U)$ of a thermodynamic system can be changed in two primary ways:
$1$. By doing work $(W)$ on the system or by the system.
$2$. By transferring heat $(Q)$ to or from the system.
According to the First Law of Thermodynamics, the change in internal energy $(\Delta U)$ is given by the equation: $\Delta U = Q - W$.
Therefore, there are $2$ ways to change the internal energy.

(N/A) The first law of thermodynamics is a statement of the law of conservation of energy. It states that if a quantity of heat $dQ$ is supplied to a system,it is used to increase the internal energy $dU$ of the system and to do work $dW$ by the system against the surroundings. Mathematically,it is expressed as: $dQ = dU + dW$.
Limitations of the first law of thermodynamics:
$1$. It does not indicate the direction of the process. For example,it does not explain why heat cannot spontaneously flow from a cold body to a hot body.
$2$. It does not specify the conditions under which the conversion of heat into work is possible.
$3$. It does not tell us about the extent of conversion of heat into work,i.e.,it does not explain why heat cannot be completely converted into work in a cyclic process.

(N/A) In thermodynamics,the sign convention for internal energy change $(\Delta U)$ is based on the First Law of Thermodynamics: $\Delta U = Q - W$.
$1$. If the internal energy of the system increases,$\Delta U$ is positive $(+ve)$. This typically occurs when heat is added to the system or work is done on the system.
$2$. If the internal energy of the system decreases,$\Delta U$ is negative $(-ve)$. This typically occurs when heat is released by the system or the system does work on the surroundings.
$3$. For an ideal gas,the internal energy depends only on temperature. Therefore,if the temperature of an ideal gas increases,$\Delta U > 0$,and if the temperature decreases,$\Delta U < 0$.

(N/A) Initially, the spring is unstretched, so the force exerted by the spring is zero. The piston is in equilibrium under the atmospheric pressure $P_a$ and the gas pressure $P_i$. Thus, $P_i = P_a$.
$(b)$ When the volume increases from $V_0$ to $V_1$, the piston moves by a distance $x = V_1 - V_0$ (since the area is unit). The spring is now stretched by $x$. The final pressure $P_f$ is balanced by atmospheric pressure, spring force, and gas pressure: $P_f = P_a + \frac{Kx}{A} = P_a + K(V_1 - V_0)$.
$(c)$ According to the first law of thermodynamics, $Q = \Delta U + W$.
For an ideal gas, $\Delta U = C_v \Delta T = \frac{C_v}{R} (P_f V_1 - P_i V_0) = \frac{f}{2} (P_f V_1 - P_i V_0)$.
The work done by the gas is $W = \int_{V_0}^{V_1} P \, dV$. Since the pressure varies linearly with volume, $W = \frac{P_i + P_f}{2} (V_1 - V_0)$.
Substituting the values: $Q = \frac{f}{2} [ (P_a + K(V_1 - V_0)) V_1 - P_a V_0 ] + \frac{P_a + P_a + K(V_1 - V_0)}{2} (V_1 - V_0)$.

(A) An extensive variable is one that depends on the size or the amount of matter in the system (e.g.,volume,mass,internal energy).
An intensive variable is one that is independent of the size or the amount of matter in the system (e.g.,pressure,temperature,density).
In the expression $P\Delta V$,$P$ (pressure) is an intensive variable,while $\Delta V$ (change in volume) is an extensive variable.
The product of an intensive variable and an extensive variable is always an extensive variable.
Therefore,$P\Delta V$ represents work done,which is an extensive property because it depends on the amount of substance in the system.

(N/A) When a gas is compressed,work is done on the gas by the external agent. According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. In a rapid compression (adiabatic process),$\Delta Q = 0$. Therefore,$\Delta U = -\Delta W$. Since work is done on the gas,$\Delta W$ is negative,which makes $\Delta U$ positive. An increase in internal energy $(\Delta U)$ leads to an increase in the temperature of the gas.

(A) Yes,it is possible to convert internal energy into work or mechanical energy.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
If a system undergoes an adiabatic process,$\Delta Q = 0$,which implies $\Delta W = -\Delta U$.
This means that any work done by the system $(\Delta W > 0)$ results in a decrease in internal energy $(-\Delta U)$.
Examples include the adiabatic expansion of a gas in a cylinder or the explosion of a bomb,where chemical internal energy is rapidly converted into mechanical kinetic energy.

(B) In a cyclic process, the system returns to its initial state. Therefore, the change in internal energy $(\Delta U)$ is zero.
According to the First Law of Thermodynamics, $\Delta Q = \Delta U + \Delta W$.
Since $\Delta U = 0$, it follows that $\Delta Q = \Delta W$.
Thus, the net amount of heat entering the system is equivalent to the net mechanical work done by the system.

(N/A) Initially,the system is in equilibrium,hence the pressure on the piston is the atmospheric pressure.
$\therefore P_{i} = P_{a}$
$(b)$ On the supply of heat,the volume of the gas increases from $V_{0}$ to $V_{1}$.
Since the cross-sectional area $A = 1$ unit,the displacement of the piston is $x = V_{1} - V_{0}$.
The force exerted by the spring on the piston is $F = kx = k(V_{1} - V_{0})$.
The final pressure $P_{f}$ on the gas is the sum of atmospheric pressure and the pressure due to the spring force:
$P_{f} = P_{a} + \frac{F}{A} = P_{a} + k(V_{1} - V_{0})$
$(c)$ According to the first law of thermodynamics,$Q = \Delta U + \Delta W$.
The change in internal energy for one mole of an ideal gas is $\Delta U = C_{V}(T_{f} - T_{0})$.
Using the ideal gas law $PV = RT$,we have $T = \frac{PV}{R}$,so $\Delta U = C_{V} \left( \frac{P_{f}V_{1}}{R} - \frac{P_{a}V_{0}}{R} \right)$.
The work done by the gas is $\Delta W = P_{a}(V_{1} - V_{0}) + \frac{1}{2}k(V_{1} - V_{0})^{2}$.
Thus,the relation is: $Q = C_{V} \left( \frac{(P_{a} + k(V_{1} - V_{0}))V_{1} - P_{a}V_{0}}{R} \right) + P_{a}(V_{1} - V_{0}) + \frac{1}{2}k(V_{1} - V_{0})^{2}$.

(A) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by the equation:
$\Delta Q = \Delta U + \Delta W$
where $\Delta Q$ is the heat supplied to the system and $\Delta W$ is the work done by the system.
First,convert the heat supplied from calories to joules:
$\Delta Q = 300 \text{ cal} \times 4.18 \text{ J/cal} = 1254 \text{ J}$.
The work done by the system is $\Delta W = 600 \text{ J}$.
Rearranging the first law equation to solve for $\Delta U$:
$\Delta U = \Delta Q - \Delta W$
$\Delta U = 1254 \text{ J} - 600 \text{ J} = 654 \text{ J}$.
Therefore,the internal energy of the system increases by $654 \text{ J}$.

(A) According to the First Law of Thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Given:
Heat supply rate $\frac{\Delta Q}{\Delta t} = \frac{6000 \, J}{60 \, sec} = 100 \, W$.
Power delivered by the system $\frac{\Delta W}{\Delta t} = 90 \, W$.
Change in internal energy $\Delta U = 2.5 \times 10^{3} \, J = 2500 \, J$.
Using the power equation: $\frac{\Delta Q}{\Delta t} = \frac{\Delta U}{\Delta t} + \frac{\Delta W}{\Delta t}$.
$100 = \frac{2500}{\Delta t} + 90$.
$100 - 90 = \frac{2500}{\Delta t}$.
$10 = \frac{2500}{\Delta t}$.
$\Delta t = \frac{2500}{10} = 250 \, sec$.

(B) For any ideal gas,the change in internal energy $\Delta U$ depends only on the change in temperature $\Delta T$ and is given by the formula:
$\Delta U = n C_V \Delta T$
For a monoatomic ideal gas,the molar heat capacity at constant volume is $C_V = \frac{3}{2} R$.
Given values are:
$n = 7 \text{ mol}$
$\Delta T = 40 K$
$R = 8.3 J K^{-1} mol^{-1}$
Substituting these values into the formula:
$\Delta U = 7 \times \left( \frac{3}{2} \times 8.3 \right) \times 40$
$\Delta U = 7 \times 3 \times 8.3 \times 20$
$\Delta U = 21 \times 166 = 3486 J$
Thus,the increase in internal energy is $3486 J$.

(D) The change in internal energy $\Delta U$ for an ideal gas is given by $\Delta U = n \cdot \frac{f}{2} \cdot R \cdot \Delta T$.
Since the container is rigid,the volume $V$ remains constant. From the ideal gas equation $pV = nRT$,we have $V \Delta p = nR \Delta T$.
Substituting this into the internal energy formula,we get $\Delta U = \frac{f}{2} (V \Delta p)$.
For a diatomic gas,the degrees of freedom $f = 5$.
Given: $V = 1 \,L = 10^{-3} \,m^{3}$ and $\Delta p = 10^{5} \,Pa$.
Calculating the value: $\Delta U = \frac{5}{2} \times 10^{-3} \times 10^{5} = 2.5 \times 10^{2} = 250 \,J$.

(B) The change in volume $\Delta V$ for $1 \,kg$ of water during phase change is given by:
$\Delta V = V_{\text{vapour}} - V_{\text{liquid}} = \frac{m}{\rho_{\text{vapour}}} - \frac{m}{\rho_{\text{liquid}}}$
$\Delta V = \frac{1}{(1/1.8)} - \frac{1}{1000} = 1.8 - 0.001 \approx 1.8 \,m^3$
Work done by the system against constant pressure $p$ is:
$W = p \Delta V = (1.01 \times 10^5 \,N m^{-2}) \times (1.8 \,m^3) = 1.818 \times 10^5 \,J$
Heat absorbed during phase change is:
$Q = mL = 1 \,kg \times 22.6 \times 10^5 \,J kg^{-1} = 22.6 \times 10^5 \,J$
Using the first law of thermodynamics,$\Delta U = Q - W$:
$\Delta U = 22.6 \times 10^5 - 1.818 \times 10^5$
$\Delta U = 20.782 \times 10^5 \,J kg^{-1} \approx 20.8 \times 10^5 \,J kg^{-1}$.

(B) The correct answer is $B$.
Internal energy is the sum of all the microscopic energies (kinetic and potential) of the molecules within a system.
The $1^{\text{st}}$ law of thermodynamics is based on the law of conservation of energy and is expressed as:
$\Delta Q = \Delta U + \Delta W$
Where:
$\Delta Q$ is the heat supplied to the system,
$\Delta U$ is the change in internal energy of the system,
$\Delta W$ is the work done by the system.
Rearranging the equation,we get $\Delta U = \Delta Q - \Delta W$. This law provides the definition and quantitative measure of the change in internal energy of a system.

(C) The first law of thermodynamics states that the change in internal energy $(dU)$ of a system is equal to the heat added to the system $(dQ)$ plus the work done on the system $(dW)$.
Mathematically,this is expressed as $dU = dQ + dW$.
If we rearrange this to solve for $dQ$,we get $dQ = dU - dW$.
Therefore,Assertion $A$ is correct.
The first law of thermodynamics is indeed a statement of the law of conservation of energy,which asserts that energy cannot be created or destroyed,only transformed. Thus,Reason $R$ is correct.
Since the mathematical expression in Assertion $A$ is derived directly from the principle of conservation of energy mentioned in Reason $R$,$R$ is the correct explanation of $A$.

(D) According to the first law of thermodynamics,the heat supplied to a system $(dQ)$ is equal to the sum of the change in internal energy $(dU)$ and the work done by the system $(dW)$: $dQ = dU + dW$.
Dividing by the time interval $(dt)$,we get the rate equation: $\frac{dQ}{dt} = \frac{dU}{dt} + \frac{dW}{dt}$.
Given:
Rate of heat supply,$\frac{dQ}{dt} = 1000 \, W$.
Rate of work done,$\frac{dW}{dt} = 200 \, W$.
Rearranging the equation to find the rate of increase of internal energy:
$\frac{dU}{dt} = \frac{dQ}{dt} - \frac{dW}{dt}$.
Substituting the values:
$\frac{dU}{dt} = 1000 \, W - 200 \, W = 800 \, W$.

(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Therefore,the change in internal energy is $\Delta U = \Delta Q - \Delta W$.
Here,$\Delta Q = mL_v = (1\,kg) \times (2257\,kJ/kg) = 2257\,kJ = 2257 \times 10^3\,J$.
The work done by the system during expansion is $\Delta W = P \Delta V = P(V_f - V_i)$.
$\Delta W = (1 \times 10^5\,Pa) \times (1.671\,m^3 - 1.00 \times 10^{-3}\,m^3) = 10^5 \times (1.671 - 0.001) = 10^5 \times 1.670 = 167000\,J = 167\,kJ$.
Now,$\Delta U = 2257\,kJ - 167\,kJ = 2090\,kJ$.

(D) According to the $1^{\text{st}}$ law of thermodynamics:
$\Delta Q = \Delta U + W$
Here,$\Delta Q = 48 \,J$,$n = 1 \,mol$,and $\Delta T = 2 \,K$ (since a change of $2^{\circ}C$ is equal to a change of $2 \,K$).
For a monatomic gas like helium,the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$.
The change in internal energy is $\Delta U = n C_v \Delta T = (1) \left(\frac{3}{2} R\right) (2) = 3R$.
Substituting the values into the first law equation:
$48 = 3R + W$
$W = 48 - 3(8.3)$
$W = 48 - 24.9$
$W = 23.1 \,J$.