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Physics
Thermodynamics
54 questions, each with its answer.
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An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 J/s, then the rate at which internal energy increases will be:From the first law of thermodynamics: $\frac{dQ}{dt} = \frac{dU}{dt} + \frac{dW}{dt}$ $100 = \frac{dU}{dt} + 75$ $\frac{dU}{dt} = 100 - 75 = 25$ W
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In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas ($\gamma=5/3$) decreases from 60 K to 50 K. The work done by the gas in the process is: (Take the universal gas constant as R = 8.3 J mol$^{-1}$ K$^{-1}$)$W=\frac{\mu R\Delta T}{1-\gamma}=\frac{1\times8.3\times(-10)}{1-\frac{5}{3}}=\frac{3\times8.3\times10}{2}=124.5$ J.
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One mole of an ideal monatomic gas undergoes a cyclic process as shown in the figure. The total heat supplied to the gas is:
For a cyclic process $\Delta U=0$, so the heat supplied equals the work done, which is the area of the rectangle: $Q=(300-100)(5-2)=200\times3=600$ J. -
A container has two chambers of volumes $V_1 = 2\text{ litres}$ and $V_2 = 3\text{ litres}$ separated by a partition made of a thermal insulator. The chambers contains $n_1 = 5$ and $n_2 = 4\text{ moles}$ of ideal gas at pressures $p_1 = 1\text{ atm}$ and $p_2 = 2\text{ atm}$, respectively. When the partition is removed, the mixture attains an equilibrium pressure of:
Total internal energy is conserved, so $P_f V_{total} = P_1 V_1 + P_2 V_2 \implies P_f = \frac{1(2) + 2(3)}{2 + 3} = \frac{8}{5} = 1.6\text{ atm}$.
For ideal gases insulated from the surroundings, internal energy is conserved upon free mixing: $U = U_1 + U_2$. Since internal energy of an ideal gas is $U = \frac{f}{2} n R T = \frac{f}{2} P V$, we have: $\frac{f}{2} P_f (V_1 + V_2) = \frac{f}{2} P_1 V_1 + \frac{f}{2} P_2 V_2 \implies P_f = \frac{P_1 V_1 + P_2 V_2}{V_1 + V_2}$. Substituting the values: $P_f = \frac{(1\text{ atm} \times 2\text{ L}) + (2\text{ atm} \times 3\text{ L})}{2\text{ L} + 3\text{ L}} = \frac{2 + 6}{5} = \frac{8}{5} = 1.6\text{ atm}$. -
Two gases A and B are filled at the same pressure in separate cylinders with movable pistons of radius $r_A$ and $r_B$, respectively. On supplying an equal amount of heat to both the systems reversibly under constant pressure, the pistons of gas A and B are displaced by $16\text{ cm}$ and $9\text{ cm}$, respectively. If the change in their internal energy is the same, then the ratio $r_A / r_B$ is equal to:
Since $Q$ and $\Delta U$ are equal, work done $W = P \Delta V$ is identical. $\pi r_A^2 (16) = \pi r_B^2 (9) \implies r_A / r_B = 3/4$.
From the first law of thermodynamics: $Q = \Delta U + W$. Since equal heat $Q_A = Q_B$ is supplied and internal energy change is identical $\Delta U_A = \Delta U_B$, the work done by both gases must be equal: $W_A = W_B$. At constant pressure $P$: $W = P \Delta V \implies P \Delta V_A = P \Delta V_B \implies \Delta V_A = \Delta V_B$. For a cylindrical piston of radius $r$ displaced by $d$: $\Delta V = (\pi r^2) d$. Thus: $\pi r_A^2 (16\text{ cm}) = \pi r_B^2 (9\text{ cm}) \implies \frac{r_A^2}{r_B^2} = \frac{9}{16} \implies \frac{r_A}{r_B} = \sqrt{\frac{9}{16}} = \frac{3}{4}$. -
A thermodynamic system is taken through the cycle $abcda$. The work done by the gas along the path $bc$ is:
Along the path bc, the volume is constant ($\Delta V = 0$). $W = \int P\,dV = 0$ So the work done along bc is zero. -
The equilibrium state of a thermodynamic system is described by A. Pressure B. Total heat C. Temperature D. Volume E. Work done. Choose the most appropriate answer from the options given below.The equilibrium state of a thermodynamic system is described by its state variables: Pressure (A), Volume (D) and Temperature (C). Total heat (B) and Work done (E) are path variables, not state variables. So the correct set is A, C and D only.
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A Carnot engine has an efficiency of 50% when its source is at a temperature $327^\circ C$. The temperature of the sink is:
$\eta=1-\frac{T_{sink}}{T_{source}}$ gives $T_{sink}=300\ K=27^\circ C$.
$\eta=1-\frac{T_{sink}}{T_{source}}=0.5\Rightarrow\frac{T_{sink}}{T_{source}}=0.5$ $\Rightarrow T_{sink}=\frac{1}{2}\times(273+327)=\frac{1}{2}\times600=300\ K=27^\circ C$ -
For the given cycle during isobaric process is:
AB is the isobaric process (constant pressure $3\times10^2\ N/m^2$, volume changing from 1 to 3 $m^3$). $W_{AB}=P(V_2-V_1)=3\times10^2\times(3-1)=3\times100\times2=600$ J. -
An ideal gas undergoes four different processes from the same initial state as shown in the figure below. Those processes are adiabatic, isothermal isobaric and isochoric. The curve which represents the adiabatic process among 1, 2, 3 and 4 is :
1: isochoric (vertical line, V constant)
2: adiabatic (steepest curve, $PV^\gamma$ = constant)
3: isothermal (less steep curve, PV = constant)
4: isobaric (horizontal line, P constant) So curve 2 represents the adiabatic process. -
An ideal gas follows a process described by the equation $PV^2=C$ from the initial $(P_1,V_1,T_1)$ to final $(P_2,V_2,T_2)$ thermodynamic states, where C is a constant. Then:$PV^2=C$ with $PV=nRT$ gives $TV=$ constant and $\frac{T^2}{P}=$ constant. So if $P_1>P_2$ then $V_2>V_1$ and $T_1>T_2$; and if $V_2>V_1$ then $T_2<T_1$.
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Two cylinders A and B of equal capacity are connected to each other via a stop cock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stop cock is suddenly opened. The process is:The entire system is thermally insulated, so no heat exchange takes place with the surroundings (Q = 0). Hence the process is adiabatic (it is a free expansion into vacuum).
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The P-V diagram for an ideal gas in a piston cylinder assembly undergoing a thermodynamic process is shown in the figure. The process is
A horizontal line on a P-V diagram represents a process at constant pressure, i.e. an isobaric process. -
The efficiency of a Carnot engine depends uponThe efficiency of a Carnot engine is $\eta=1-\dfrac{T_2}{T_1}$, where $T_1$ is the source temperature and $T_2$ is the sink temperature — it depends on both temperatures.
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In which of the following processes, heat is neither absorbed nor released by a system?In an adiabatic process there is no exchange of heat between the system and the surroundings, i.e. $\Delta Q=0$. Hence heat is neither absorbed nor released in an adiabatic process.
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1 g of water, of volume 1 cm$^3$ at $100^\circ C$, is converted into steam at same temperature under normal atmospheric pressure ($=1\times10^5$ Pa). The volume of steam formed equals 1671 cm$^3$. If the specific latent heat of vaporisation of water is 2256 J/g, the change in internal energy is,$Q=\Delta U+P\Delta V \Rightarrow 1\times2256=\Delta U+10^5\times(1670\times10^{-6}) \Rightarrow \Delta U=2256-167=2089$ J.
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The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is
For an isobaric process $W/Q = R/C_p = 2/5$ for a monatomic gas.
The V–T graph is a straight line through the origin, so the given process is isobaric. $dQ = nC_p\,dT = n\left(\frac{5}{2}R\right)dT$ (monatomic gas) $dW = P\,dV = nR\,dT$ Required ratio $= \frac{dW}{dQ} = \frac{nR\,dT}{n\left(\frac{5}{2}R\right)dT} = \frac{2}{5}$ So the ratio of work done to heat absorbed is $\frac{2}{5}$. -
The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is
$\eta = 1 - T_2/T_1$ with $T_1 = 373$ K and $T_2 = 273$ K.
Efficiency of ideal heat engine, $\eta = \left(1 - \frac{T_2}{T_1}\right)$ $T_2$ : sink temperature, $T_1$ : source temperature $\%\eta = \left(1 - \frac{T_2}{T_1}\right)\times 100 = \left(1 - \frac{273}{373}\right)\times 100$ $= \left(\frac{100}{373}\right)\times 100 = 26.8\%$ -
A sample of 0.1 g of water at $100^\circ C$ and normal pressure ($1.013 \times 10^{5}\ N m^{-2}$) requires 54 cal of heat energy to convert to steam at $100^\circ C$. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is
$\Delta U = Q - P\Delta V$.
$\Delta Q = \Delta U + \Delta W$ $54\times4.18 = \Delta U + 1.013\times10^{5}(167.1\times10^{-6} - 0)$ $\Delta U = 225.7 - 16.9 = 208.7$ J -
Thermodynamic processes are indicated in the following diagram. Match the following
Pick the matching item from Column II for each row.
Column I
- P. Process I
- Q. Process II
- R. Process III
- S. Process IV
Column II
- a. Adiabatic
- b. Isobaric
- c. Isochoric
- d. Isothermal
Correct answer: P → c, Q → a, R → d, S → b
Vertical line = isochoric, horizontal = isobaric, along isotherm = isothermal, steeper curve = adiabatic.
Process I = Isochoric Process II = Adiabatic Process III = Isothermal Process IV = Isobaric Correct match: P-c, Q-a, R-d, S-b. -
A Carnot engine having an efficiency of $\frac{1}{10}$ as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is
COP of the reversed engine is $(1-\eta)/\eta = 9$.
Coefficient of performance $\beta = \frac{1 - \eta}{\eta} = \frac{1 - \frac{1}{10}}{\frac{1}{10}} = 9$ $\beta = \frac{Q_2}{W}$ $Q_2 = 9\times10 = 90$ J -
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then
The adiabatic curve is steeper, enclosing more area on compression.
Isothermal compression from V to $\frac{V}{2}$: $W = nRT\ln\left(\frac{V_f}{V_i}\right) = nRT\ln\left(\frac{1}{2}\right)$
Adiabatic compression: $W = \frac{k(V_f^{1-\gamma} - V_i^{1-\gamma})}{1 - \gamma}$, with $PV^\gamma$ = constant
The area under the adiabatic compression curve is greater than that under the isothermal compression curve.
Hence compressing the gas through the adiabatic process requires more work. -
A refrigerator works between 4°C and 30°C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is [Take, 1 cal = 4.2 J]'Out of Syllabus'
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One mole of an ideal monatomic gas undergoes a process described by the equation $PV^3$ = constant. The heat capacity of the gas during this process is
$C = C_V + R/(1 - x)$.
$PV^x$ = constant (polytropic process); heat capacity $C = C_V + \frac{R}{1 - x}$
Given x = 3 and the gas is monatomic (f = 3)
$C = \frac{fR}{2} + \frac{R}{1 - x} = \frac{3}{2}R - \frac{R}{2} = R$ -
The temperature inside a refrigerator is $t_2$°C and the room temperature is $t_1$°C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be:
$Q_1/W = T_1/(T_1 - T_2)$.
Heat delivered to the room $= Q_1$
COP $\beta = \frac{t_2 + 273}{t_1 - t_2} = \frac{Q_2}{W} = \frac{Q_1 - W}{W} = \frac{Q_1}{W} - 1$
$\Rightarrow \frac{Q_1}{W} = 1 + \frac{t_2 + 273}{t_1 - t_2} = \frac{t_1 + 273}{t_1 - t_2}$
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