First Law of Thermodynamics Questions in English

(B) The mechanical equivalent of heat is given by $J = 4.18 \text{ J/cal}$.
Given the heat supplied $Q = 400 \text{ calories}$.
The work done $W$ is calculated using the relation $W = J \times Q$.
Substituting the values,we get $W = 4.18 \times 400 = 1672 \text{ Joules}$.

(A) The first law of thermodynamics is a statement of the conservation of energy applied to thermodynamic systems. It states that the heat supplied to a system $(dQ)$ is equal to the change in internal energy $(dU)$ plus the work done by the system $(dW)$.
Mathematically,it is expressed as: $dQ = dU + dW$.
Since the work done by a gas during an expansion or compression is given by $dW = PdV$,we substitute this into the equation.
Therefore,the first law of thermodynamics is given by: $dQ = dU + PdV$.

(B) According to the First Law of Thermodynamics, the heat supplied to a system ($\Delta Q$) is equal to the sum of the change in internal energy ($\Delta U$) and the work done by the system ($\Delta W$).
Mathematically, this is expressed as: $\Delta Q = \Delta U + \Delta W$.
Given that the heat supplied is $Q$ and the work done by the system is $W$, we substitute these values into the equation: $Q = \Delta U + W$.
Rearranging the equation to solve for the change in internal energy ($\Delta U$): $\Delta U = Q - W$.

(B) According to the First Law of Thermodynamics,the change in internal energy $(\Delta U)$ is given by the equation: $\Delta U = \Delta Q - W$.
Here,the heat given to the system $(\Delta Q)$ is $+35 \ J$ and the work done by the system $(W)$ is $+15 \ J$.
Substituting these values into the equation:
$\Delta U = 35 \ J - 15 \ J = 20 \ J$.
Therefore,the change in the internal energy of the system is $20 \ J$.

(B) According to the first law of thermodynamics,
$Q = \Delta U + W$
where $Q$ is the heat supplied to the system,$\Delta U$ is the change in internal energy,and $W$ is the work done by the system.
This law is a statement of the law of conservation of energy,which states that energy can neither be created nor destroyed,only transformed from one form to another.

(D) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by the equation: $\Delta Q = \Delta U + \Delta W$.
Here,the heat given to the system is $\Delta Q = 35 \ J$.
The work done by the system is $\Delta W = -15 \ J$.
Substituting these values into the equation: $35 = \Delta U + (-15)$.
Solving for $\Delta U$: $\Delta U = 35 + 15 = 50 \ J$.
Therefore,the change in internal energy is $50 \ J$.

(A) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by the equation: $\Delta U = \Delta Q - \Delta W$.
Here,the heat supplied $\Delta Q = 300 \text{ calories}$.
Converting heat into joules: $\Delta Q = 300 \times 4.18 \text{ J} = 1254 \text{ J}$.
The work done by the system $\Delta W = 600 \text{ J}$.
Substituting these values into the equation: $\Delta U = 1254 \text{ J} - 600 \text{ J} = 654 \text{ J}$.
Therefore,the internal energy of the system increases by $654 \text{ J}$.

(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta Q$ is the heat absorbed,$\Delta U$ is the change in internal energy,and $\Delta W$ is the work done by the system.
Given:
Heat absorbed,$\Delta Q = 2 \; kcal = 2 \times 10^3 \times 4.2 \; J = 8400 \; J$.
Work done,$\Delta W = 500 \; J$.
Substituting these values into the equation:
$\Delta U = \Delta Q - \Delta W$
$\Delta U = 8400 \; J - 500 \; J = 7900 \; J$.
Therefore,the internal energy change is $7900 \; J$.

(D) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
In this equation,$\Delta Q$ (heat) and $\Delta W$ (work) are path-dependent quantities,meaning their values depend on the process or path taken to reach the final state from the initial state.
However,$\Delta U$ (change in internal energy) is a state function.
$A$ state function is a property whose value depends only on the current state of the system (initial and final states) and is independent of the path taken to reach that state.
Therefore,$\Delta U$ is the unique function of the initial and final states.

(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.
$\Delta Q = \Delta U + \Delta W$
Given:
Heat added,$\Delta Q = 110\; J$
Change in internal energy,$\Delta U = 40\; J$
Rearranging the formula to find the work done:
$\Delta W = \Delta Q - \Delta U$
$\Delta W = 110\; J - 40\; J = 70\; J$
Therefore,the amount of external work done is $70\; J$.

(B) thermodynamic state function is a property whose value depends only on the current state of the system and not on the path taken to reach that state.
Enthalpy $(H)$,Gibbs energy $(G)$,and Internal energy $(U)$ are all state functions because they depend only on the state variables of the system.
Work done $(W)$ is a path function,meaning its value depends on the specific process or path taken to change the state of the system.
Therefore,work done is not a thermodynamic state function.

(C) According to the First Law of Thermodynamics,the heat supplied to a system $(\Delta Q)$ is equal to the sum of the change in internal energy $(\Delta U)$ and the work done by the system $(\Delta W)$.
Formula: $\Delta Q = \Delta U + \Delta W$
Given:
$\Delta W = 333 \ cal$
$\Delta U = 167 \ cal$
Calculation:
$\Delta Q = 167 \ cal + 333 \ cal = 500 \ cal$
Therefore,the heat supplied is $500 \ cal$.

(D) The first law of thermodynamics is essentially the law of conservation of energy applied to thermodynamic systems. It states that energy can neither be created nor destroyed,only transformed from one form to another. Since heat is a form of energy,the first law implies that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system $(ΔU = Q - W)$. Therefore,heat is recognized as a form of energy in transit.

(D) According to the First Law of Thermodynamics,the change in internal energy $\Delta U$ is given by $\Delta U = \Delta Q - \Delta W$.
Internal energy is a state function,meaning its value depends only on the initial and final states of the system and is independent of the path taken.
Since the initial state $(P_1, V_1)$ and the final state $(P_2, V_2)$ are the same for both processes,the change in internal energy $\Delta U$ will remain the same.
Therefore,the quantity $\Delta Q - \Delta W$ remains constant.

(B) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by the equation: $\Delta Q = \Delta U + \Delta W$.
Here,heat is given to the system,so $\Delta Q = +200 \ J$.
Work is done on the system,so $\Delta W = -100 \ J$.
Substituting these values into the equation: $200 = \Delta U + (-100)$.
Therefore,$\Delta U = 200 + 100 = 300 \ J$.

(D) According to the First Law of Thermodynamics,the change in internal energy $(\Delta U)$ is given by the equation: $\Delta Q = \Delta U + \Delta W$.
Here,the heat added to the system is $\Delta Q = 150 \ J$.
The work done by the system is $\Delta W = 110 \ J$.
Rearranging the formula to solve for $\Delta U$: $\Delta U = \Delta Q - \Delta W$.
Substituting the values: $\Delta U = 150 \ J - 110 \ J = 40 \ J$.
Therefore,the change in internal energy is $40 \ J$.

(B) The first law of thermodynamics is generally expressed as $\Delta Q = \Delta U + W_{by}$,where $W_{by}$ is the work done $BY$ the system.
In this problem,$\Delta W$ is defined as the work done $ON$ the system.
By sign convention,work done on the system is negative of work done by the system,so $W_{by} = -\Delta W$.
Substituting this into the first law equation: $\Delta Q = \Delta U + (-\Delta W)$.
Therefore,the correct expression is $\Delta Q = \Delta U - \Delta W$.

(C) According to the First Law of Thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Given,heat supplied to the system $\Delta Q = 200 \, cal$.
Converting calories to Joules,$\Delta Q = 200 \times 4.2 \, J = 840 \, J$.
Work done by the system $\Delta W = 40 \, J$.
Substituting these values into the equation,$\Delta U = \Delta Q - \Delta W$.
$\Delta U = 840 \, J - 40 \, J = 800 \, J$.
Since $\Delta U$ is positive,the internal energy increases by $800 \, J$.

(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Given:
Heat released by the gas,$\Delta Q = -20 \ J$ (heat lost by the system is negative).
Work done on the gas,$\Delta W = -10 \ J$ (work done on the system is negative in the convention $\Delta Q = \Delta U + \Delta W$).
Initial internal energy,$U_i = 40 \ J$.
Substituting the values into the equation:
$-20 = (U_f - 40) + (-10)$
$-20 = U_f - 40 - 10$
$-20 = U_f - 50$
$U_f = 50 - 20 = 30 \ J$.
Wait,re-evaluating the sign convention: If $\Delta Q = \Delta U + W$ where $W$ is work done $BY$ the gas,then $W = -10 \ J$. Thus,$-20 = \Delta U - 10$,so $\Delta U = -10 \ J$. Since $\Delta U = U_f - U_i$,we have $U_f - 40 = -10$,which gives $U_f = 30 \ J$.

(B) The first law of thermodynamics states that the heat $Q$ supplied to a system is equal to the change in internal energy $\Delta U$ plus the work done $W$ by the system,expressed as $Q = \Delta U + W$.
This equation is a direct application of the law of conservation of energy,which states that energy cannot be created or destroyed,only transformed from one form to another.
Therefore,the first law of thermodynamics is a special case of the law of conservation of energy.

(C) According to the First Law of Thermodynamics $(FLOT)$,the heat supplied $(\Delta Q)$ is equal to the change in internal energy $(\Delta U)$ plus the work done by the gas $(W = P \Delta V)$.
$\Delta Q = \Delta U + P \Delta V$
Given:
$\Delta Q = 1500 \; J$
$P = 2.1 \times 10^5 \; N/m^2$
$\Delta V = 2.5 \times 10^{-3} \; m^3$
Rearranging the formula to find $\Delta U$:
$\Delta U = \Delta Q - P \Delta V$
$\Delta U = 1500 - (2.1 \times 10^5) \times (2.5 \times 10^{-3})$
$\Delta U = 1500 - (2.1 \times 2.5 \times 10^2)$
$\Delta U = 1500 - (5.25 \times 100)$
$\Delta U = 1500 - 525$
$\Delta U = 975 \; J$
Therefore,the increase in internal energy is $975 \; J$.

(A) 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.
Given: $\Delta Q = 6 \, kcal = 6 \times 4.184 \, kJ = 25.104 \, kJ$ (using $1 \, kcal = 4.184 \, kJ$).
Work done $\Delta W = 6 \, kJ$.
Therefore,$\Delta U = \Delta Q - \Delta W = 25.104 \, kJ - 6 \, kJ = 19.104 \, kJ$.
Rounding to one decimal place,we get $\Delta U \approx 19.1 \, kJ$.

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

(C) The change in internal energy of an ideal gas is given by the formula $\Delta U = n C_V \Delta T$.
Since the gas is monoatomic,the molar heat capacity at constant volume $C_V$ is constant,given by $C_V = \frac{3}{2}R$.
Internal energy is a state function,meaning it depends only on the initial and final states of the system (temperature $T_1$ and $T_2$) and not on the path or process taken to reach the final state.
Therefore,for a given change in temperature $\Delta T = T_2 - T_1$,the change in internal energy $\Delta U$ remains the same regardless of whether the process occurs at constant volume or constant pressure.

(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
In the first process:
$\Delta Q_1 = 8 \times 10^5 \ J$ and $\Delta W_1 = 6.5 \times 10^5 \ J$.
Therefore,$\Delta U = \Delta Q_1 - \Delta W_1 = (8 - 6.5) \times 10^5 \ J = 1.5 \times 10^5 \ J$.
Since the initial and final states are the same for both processes,the change in internal energy $\Delta U$ remains constant.
In the second process:
$\Delta Q_2 = 10^5 \ J$ and $\Delta U = 1.5 \times 10^5 \ J$.
Using $\Delta Q_2 = \Delta U + \Delta W_2$:
$10^5 = 1.5 \times 10^5 + \Delta W_2$.
$\Delta W_2 = 10^5 - 1.5 \times 10^5 = -0.5 \times 10^5 \ J$.
$A$ negative value for work done by the gas indicates that work is done on the gas. Thus,the work done on the gas is $0.5 \times 10^5 \ J$.

(C) The work done by a system is given by the formula $\Delta W = P \Delta V$,where $P$ is the pressure and $\Delta V$ is the change in volume.
Since the system undergoes contraction,the final volume is less than the initial volume,which means $\Delta V = V_{final} - V_{initial} < 0$.
Therefore,the work done by the system $\Delta W = P \Delta V$ will be negative.

(B) The first law of thermodynamics is based on the law of conservation of energy and introduces the concept of internal energy $(U)$. It states that $\Delta Q = \Delta U + \Delta W$. It is applicable to all processes,including cyclic processes. However,the concept of entropy is introduced by the second law of thermodynamics,not the first. Therefore,statement $(b)$ is incorrect.

(C) According to the first law of thermodynamics,the heat supplied $(\Delta Q)$ is equal to the sum of the change in internal energy $(\Delta U)$ and the work done $(\Delta W)$:
$\Delta Q = \Delta U + \Delta W$
Given:
Latent heat of vaporisation $(\Delta Q)$ = $2240 \, J$
Work done $(\Delta W)$ = $168 \, J$
Rearranging the formula to find the change in internal energy:
$\Delta U = \Delta Q - \Delta W$
$\Delta U = 2240 \, J - 168 \, J$
$\Delta U = 2072 \, J$
Therefore,the increase in internal energy is $2072 \, J$.

(B) Given: Heat supplied $Q = 540 \; cal$.
Initial volume of water $V_1 = 1 \; cm^3$.
Final volume of steam $V_2 = 1671 \; cm^3$.
Change in volume $\Delta V = V_2 - V_1 = 1671 - 1 = 1670 \; cm^3$.
Atmospheric pressure $P = 1.013 \times 10^5 \; Pa = 1.013 \times 10^6 \; dyne/cm^2$.
Work done against atmospheric pressure $W = P \Delta V$.
$W = (1.013 \times 10^6 \; dyne/cm^2) \times (1670 \; cm^3) = 1.69171 \times 10^9 \; ergs$.
Since $1 \; cal = 4.184 \times 10^7 \; ergs$,we convert the work into calories:
$W = \frac{1.69171 \times 10^9}{4.184 \times 10^7} \approx 40.4 \; cal$.
Thus,the work done is nearly $40 \; cal$.

(C) According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.
Here,$\Delta Q$ is the heat supplied,which is equal to the latent heat $L$ for a unit mass.
$W$ is the work done by the system during expansion at constant pressure $P$,given by $W = P(V_2 - V_1)$.
Substituting these into the first law equation:
$L = \Delta U + P(V_2 - V_1)$.
Rearranging for the change in internal energy $\Delta U$:
$\Delta U = L - P(V_2 - V_1)$.

(C) According to the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$.
Here, $\Delta Q$ is the total heat absorbed, $\Delta U$ is the change in internal energy (energy spent in overcoming intermolecular forces), and $\Delta W$ is the work done against atmospheric pressure.
Given: $\Delta Q = 540 \text{ cal}$, $P = 1.013 \times 10^5 \text{ N/m}^2$, $V_1 = 1 \text{ cm}^3 = 10^{-6} \text{ m}^3$, $V_2 = 1671 \text{ cm}^3 = 1671 \times 10^{-6} \text{ m}^3$, and $J = 4.19 \text{ J/cal}$.
Work done $\Delta W = P(V_2 - V_1) = 1.013 \times 10^5 \times (1671 - 1) \times 10^{-6} \text{ J} = 1.013 \times 10^5 \times 1670 \times 10^{-6} \text{ J} \approx 169.17 \text{ J}$.
Converting work to calories: $\Delta W_{\text{cal}} = \frac{169.17}{4.19} \approx 40.37 \text{ cal}$.
Internal energy change $\Delta U = \Delta Q - \Delta W_{\text{cal}} = 540 - 40.37 \approx 500 \text{ cal}$.

(D) The container is rigid,which means the volume of the ideal gas remains constant,so the work done $W = P\Delta V = 0$.
According to the First Law of Thermodynamics,$\Delta Q = \Delta U + W$.
Since $W = 0$,the change in internal energy is equal to the heat supplied: $\Delta U = \Delta Q$.
The heat supplied by the filament is given by Joule's heating formula: $\Delta Q = I^2Rt$.
Given: $I = 1 \,A$,$R = 100 \,\Omega$,and $t = 5 \,min = 5 \times 60 \,s = 300 \,s$.
Substituting the values: $\Delta U = (1)^2 \times 100 \times 300 = 30,000 \,J$.
Converting to $kJ$: $\Delta U = 30 \,kJ$.

(D) Given,heat supplied to the system,$\Delta Q = 20 \, cal = 20 \times 4.2 \, J = 84 \, J$.
Work done on the system,$\Delta W = -50 \, J$ (Since work is done on the system,it is negative).
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta U$ is the change in internal energy.
Therefore,$\Delta U = \Delta Q - \Delta W$.
Substituting the values,$\Delta U = 84 \, J - (-50 \, J) = 84 \, J + 50 \, J = 134 \, J$.
Thus,the change in internal energy between $A$ and $C$ is $134 \, J$.

(D) The change in internal energy $\Delta U$ is a state function,meaning it depends only on the initial and final states,not on the path taken.
For the path $iaf$,the first law of thermodynamics is given by $\Delta Q = \Delta U + \Delta W$.
Given $\Delta Q_{iaf} = 50 \, J$ and $\Delta W_{iaf} = 20 \, J$,we have:
$\Delta U_{if} = \Delta Q_{iaf} - \Delta W_{iaf} = 50 \, J - 20 \, J = 30 \, J$.
For the return path $fi$,the initial state is $f$ and the final state is $i$. Therefore,the change in internal energy is $\Delta U_{fi} = -\Delta U_{if} = -30 \, J$.
Using the first law of thermodynamics for the path $fi$:
$\Delta Q_{fi} = \Delta U_{fi} + \Delta W_{fi}$.
Given $\Delta W_{fi} = -13 \, J$ and $\Delta U_{fi} = -30 \, J$,we get:
$\Delta Q_{fi} = -30 \, J + (-13 \, J) = -43 \, J$.

(A) According to the first law of thermodynamics,the heat supplied to a system $(\Delta Q)$ is equal to the sum of the change in internal energy $(\Delta U)$ and the work done by the system $(\Delta W)$:
$\Delta Q = \Delta U + \Delta W$
Given:
Heat supplied,$\Delta Q = 40 \ J$
Work done,$\Delta W = 30 \ J$
Substituting these values into the equation:
$40 \ J = \Delta U + 30 \ J$
$\Delta U = 40 \ J - 30 \ J = 10 \ J$
Therefore,the change in internal energy between $A$ and $C$ is $10 \ J$.

(A) From the given $P-V$ diagram,process $AB$ is an isochoric process (volume is constant).
For process $AB$:
$\Delta W_{AB} = 0$
Using the first law of thermodynamics,$\Delta Q_{AB} = \Delta U_{AB} + \Delta W_{AB}$.
Given $\Delta Q_{AB} = 50 \ J$,so $\Delta U_{AB} = 50 \ J$.
Since $U_A = 1500 \ J$,then $U_B = U_A + \Delta U_{AB} = 1500 + 50 = 1550 \ J$.
For process $BC$:
Given $\Delta Q_{BC} = 0$ (adiabatic process).
Work done on the gas is $\Delta W_{BC} = -40 \ J$ (since work is done on the gas,it is negative).
Using the first law,$\Delta Q_{BC} = \Delta U_{BC} + \Delta W_{BC}$.
$0 = \Delta U_{BC} - 40 \ J \implies \Delta U_{BC} = 40 \ J$.
Therefore,$U_C = U_B + \Delta U_{BC} = 1550 + 40 = 1590 \ J$.

(B) The first law of thermodynamics states that $\Delta Q = \Delta U + \Delta W$. This equation represents the principle of conservation of energy,where the heat supplied to a system $(\Delta Q)$ is used to increase its internal energy $(\Delta U)$ and to perform work against external pressure $(\Delta W)$. Thus,it is a direct application of the law of conservation of energy.

(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Here,the work done on the gas during compression is $\Delta W = P \Delta V$.
Since the volume changes from $10 \ m^{3}$ to $4 \ m^{3}$,$\Delta V = (4 - 10) \ m^{3} = -6 \ m^{3}$.
Thus,$\Delta W = 50 \times (-6) = -300 \ J$.
The heat supplied is $\Delta Q = 100 \ J$.
Using the first law: $\Delta U = \Delta Q - \Delta W$.
$\Delta U = 100 - (-300) = 100 + 300 = 400 \ J$.

(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Here,the gas releases heat,so $\Delta Q = -20 \ J$.
Work is done on the gas,so $\Delta W = -10 \ J$.
The initial internal energy $U_1 = 40 \ J$.
Substituting these values into the equation: $-20 = (U_2 - 40) + (-10)$.
$-20 = U_2 - 40 - 10$.
$-20 = U_2 - 50$.
$U_2 = 50 - 20 = 30 \ J$.

(D) Since the container is rigid,the volume of the ideal gas remains constant,so the work done $W = P \Delta V = 0$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + W$. Since $W = 0$,we have $\Delta Q = \Delta U$.
The heat supplied by the filament is given by Joule's law of heating: $\Delta Q = i^2 R t$.
Given: $i = 1 \, A$,$R = 100 \, \Omega$,and $t = 5 \, \text{min} = 5 \times 60 \, \text{s} = 300 \, \text{s}$.
Substituting the values: $\Delta U = (1)^2 \times 100 \times 300 = 30,000 \, \text{J} = 30 \, \text{kJ}$.

(C) According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.
Here,the work done by the gas at constant pressure is $W = P \Delta V$.
Given: $\Delta Q = 1500 \ J$,$P = 2.5 \times 10^{5} \ N/m^{2}$,and $\Delta V = 2.5 \times 10^{-3} \ m^{3}$.
Calculating the work done: $W = (2.5 \times 10^{5}) \times (2.5 \times 10^{-3}) = 6.25 \times 10^{2} = 625 \ J$.
Now,substituting the values into the first law equation: $\Delta U = \Delta Q - W$.
$\Delta U = 1500 \ J - 625 \ J = 875 \ J$.
Therefore,the increase in internal energy is $875 \ J$.

(D) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,which implies $\Delta U = \Delta Q - \Delta W$.
In this problem,the heat supplied to the system is $\Delta Q = +35 \ J$.
Since work is done on the system,the work done by the system is negative,so $\Delta W = -15 \ J$.
Substituting these values into the equation: $\Delta U = 35 \ J - (-15 \ J) = 35 \ J + 15 \ J = 50 \ J$.
Therefore,the change in internal energy is $50 \ J$.

(D) The work done by the system is given by the formula $W = P \Delta V$.
Here,the atmospheric pressure $P = 1.013 \times 10^5 \ Pa$.
The change in volume is $\Delta V = V_2 - V_1 = 3.34 \ m^3 - 0.002 \ m^3 = 3.338 \ m^3$.
Substituting the values: $W = (1.013 \times 10^5 \ Pa) \times (3.338 \ m^3)$.
$W \approx 3.38 \times 10^5 \ J = 338 \ kJ$.
Rounding to the nearest provided option,the work done is approximately $340 \ kJ$.

(D) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Here,the heat absorbed $\Delta Q = 2 \, kcal = 2 \times 10^3 \, cal$.
Since $1 \, cal = 4.2 \, J$,we have $\Delta Q = 2000 \times 4.2 = 8400 \, J$.
The work done by the system is $\Delta W = 500 \, J$.
Using the formula $\Delta U = \Delta Q - \Delta W$,we get:
$\Delta U = 8400 \, J - 500 \, J = 7900 \, J$.
Therefore,the change in internal energy is $7900 \, J$.

(C) According to the first law of thermodynamics:
$\Delta Q = \Delta U + \Delta W$
Since the process occurs at constant pressure,the work done is $\Delta W = P \Delta V$.
Therefore,$\Delta Q = \Delta U + P \Delta V$.
Given:
$\Delta Q = 1500 \; J$
$P = 2.1 \times 10^{5} \; N/m^{2}$
$\Delta V = 2.5 \times 10^{-3} \; m^{3}$
Calculating the work done:
$\Delta W = P \Delta V = (2.1 \times 10^{5}) \times (2.5 \times 10^{-3}) = 525 \; J$.
Now,calculating the change in internal energy:
$\Delta U = \Delta Q - \Delta W = 1500 - 525 = 975 \; J$.

(B) The kinetic energy of the container is $K = \frac{1}{2} M v^2$ (since it contains $1 \text{ mole}$,mass $m = M$).
When the container stops,this kinetic energy is converted into the internal energy of the gas,$\Delta U = \mu C_V \Delta T$.
Since $\mu = 1 \text{ mole}$,we have $\Delta U = C_V \Delta T$.
We know that $C_V = \frac{R}{\gamma - 1}$.
Equating the kinetic energy to the change in internal energy:
$\frac{1}{2} M v^2 = C_V \Delta T$
$\frac{1}{2} M v^2 = \frac{R}{\gamma - 1} \Delta T$
Solving for $\Delta T$:
$\Delta T = \frac{M v^2 (\gamma - 1)}{2R}$.

(B) The change in internal energy $\Delta U$ for an ideal gas is given by the formula: $\Delta U = \mu C_v \Delta T$.
Given:
Number of moles $\mu = 2 \ mol$.
Specific heat at constant volume $C_v = 4.96 \ cal/mole \ K$.
Change in temperature $\Delta T = 342 \ K - 340 \ K = 2 \ K$.
Substituting these values into the formula:
$\Delta U = 2 \times 4.96 \times 2$.
$\Delta U = 19.84 \ cal$.

(D) According to the first law of thermodynamics,the heat supplied to a system $(\Delta Q)$ is equal to the sum of the change in internal energy $(\Delta U)$ and the work done by the system $(\Delta W)$:
$\Delta Q = \Delta U + \Delta W$
Given:
Heat supplied,$\Delta Q = 150 \ J$
Work done by the system,$\Delta W = 110 \ J$
Substituting these values into the equation:
$150 = \Delta U + 110$
$\Delta U = 150 - 110$
$\Delta U = 40 \ J$
Therefore,the change in internal energy is $40 \ J$.

(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.
$\Delta Q = \Delta U + \Delta W$
Given:
$\Delta Q = 110 \ J$
$\Delta U = 40 \ J$
Substituting the values into the equation:
$110 = 40 + \Delta W$
$\Delta W = 110 - 40$
$\Delta W = 70 \ J$
Therefore,the work done is $70 \ J$.