Thermodynamics — NEET Physics PYQ MCQs with Solutions

Solution: Given $m=1$ g, volume of water $=1\,\text{cm}^3=10^{-6}\,\text{m}^3$, volume of steam $=1671\times10^{-6}\,\text{m}^3$, $p=1\times10^5$ Pa, $L=2256$ J/g. Change in volume $\Delta V=(1671-1)\times10^{-6}=1670\times10^{-6}\,\text{m}^3$. Heat supplied $\Delta Q=mL=1\times2256=2256$ J. Work done in expansion $\Delta W=p\Delta V=1\times10^5\times1670\times10^{-6}=167$ J. By the first law, $\Delta U=\Delta Q-\Delta W=2256-167=2089$ J.

Solution: Heat spent, $\Delta Q=54$ cal $=54\times4.18=225.72$ J. Normal pressure $p=1.013\times10^5\,\text{Nm}^{-2}$. Net work done $\Delta W=p\Delta V=p[V_{\text{steam}}-V_{\text{water}}]$ with $V_{\text{steam}}=167.1\times10^{-6}\,\text{m}^3$ and $V_{\text{water}}=0.1\times10^{-6}\,\text{m}^3$. So $\Delta W=1.013\times10^5[(167.1-0.1)\times10^{-6}]=1.013\times167\times10^{-1}=16.917$ J. By the first law, $\Delta U=\Delta Q-\Delta W=225.72-16.917=208.7$ J.

Solution: By the first law, $\Delta U=\Delta Q+\Delta W$. For an isothermal process $\Delta U=0$, so $\Delta Q=-\Delta W$. Given $\Delta W=-150$ J, thus $\Delta Q=+150$ J. Since $Q$ is positive, heat has been added to the gas.

Solution: From the first law of thermodynamics $\Delta Q=\Delta U+\Delta W$. For an adiabatic process $\Delta Q=0$, therefore $\Delta U=-\Delta W$.

Solution: By the first law, $\Delta U=Q-W$, where $Q$ is heat given to the system and $W$ the work done. $\Delta U=2\times4.2\times1000-500=8400-500=7900$ J.

Solution: For a cyclic process, the internal energy returns to its initial value, so $\Delta U=0$, i.e. $E=0$.

Solution: An isothermal process is at constant temperature, so $\Delta T=0$ and $\Delta U=nC_v\Delta T=0$, giving $\Delta Q=\Delta W$. An adiabatic process has no heat transfer, so $\Delta Q=0$; from the first law $\Delta Q=\Delta U+\Delta W$ this gives $\Delta U=-W$.

Solution: By the first law $\Delta Q=\Delta U+\Delta W$, with $\Delta Q=110$ J and $\Delta U=40$ J. Therefore $\Delta W=\Delta Q-\Delta U=110-40=70$ J.

Solution: By the first law, the heat supplied $dQ$ equals the increase in internal energy $dU$ plus the external work done $dW$, i.e. $dQ=dU+dW$. This is essentially the law of conservation of energy applied to every process in nature.

Solution: From the $p$-$V$ diagram, the pressure of the ideal gas in the piston cylinder is constant during the thermodynamic process. Hence, this process is isobaric.

Solution: Since the entire system is thermally insulated, there is no transfer of heat between the system and the surroundings. When the stop cock is suddenly opened, a sudden expansion takes place with no heat transfer. Such a process, occurring without transfer of heat or mass, is adiabatic.

Solution: In an adiabatic process the system is completely insulated from the surroundings, so heat is neither absorbed nor released, $\Delta Q=0$. Sudden processes such as the bursting of a cycle tyre are adiabatic. (Isobaric: $\Delta p=0$; isothermal: $\Delta T=0$; isochoric: $\Delta V=0$.)

Solution: From the graph $V\propto T$, so $\dfrac{V}{T}=$ constant, i.e. the process is isobaric. Work done $\Delta W=p\Delta V=nR\Delta T=nR(T_B-T_A)$ and heat absorbed $\Delta Q=nC_p\Delta T=nC_p(T_B-T_A)$, with $C_p=\dfrac{\gamma R}{\gamma-1}$ and $\gamma=1+\dfrac{2}{f}$. For a monoatomic gas $f=3$, so $C_p=\dfrac{5}{2}R$. Hence $\dfrac{\Delta W}{\Delta Q}=\dfrac{nR(T_B-T_A)}{n\left(\frac{5}{2}R\right)(T_B-T_A)}=\dfrac{2}{5}$.

Solution: In an isochoric process the curve is parallel to the $p$-axis (volume constant); in an isobaric process it is parallel to the $V$-axis (pressure constant); along an isotherm the temperature is constant. So $P\to c$, $Q\to a$, $R\to d$ and $S\to b$.

Solution: Plotting the $p$-$V$ graph for isothermal and adiabatic compression of the gas to half its initial volume, the adiabatic curve is steeper. Since work done equals the area under the $p$-$V$ curve, $W_{adiabatic}>W_{isothermal}>W_{isobaric}$, so compressing through the adiabatic process requires more work.

Solution: Since the initial and final points are the same, $\Delta U_{A\to B\to C}=\Delta U_{A\to C}$. $A\to B$ is isochoric, so $dW_{A\to B}=0$ and $dQ_{A\to B}=dU_{A\to B}=400$ J. $B\to C$ is isobaric, so $dQ_{B\to C}=dU_{B\to C}+p\Delta V_{B\to C}$: $100=dU_{B\to C}+6\times10^4(2\times10^{-3})$, giving $dU_{B\to C}=100-120=-20$ J. Then $dQ_{A\to C}=380+\left(2\times10^4\times2\times10^{-3}+\dfrac{1}{2}\times2\times10^{-3}\times4\times10^4\right)=380+(40+40)=460$ J.

Solution: For a diatomic gas, $C_V=\dfrac{5}{2}R$. $\Delta U=nC_V\,dT=n\dfrac{5R}{2}(T_B-T_A)=\dfrac{5}{2}(p_BV_B-p_AV_A)=\dfrac{5}{2}(2\times10^{3}\times6-5\times10^{3}\times4)=\dfrac{5}{2}\times(-8\times10^{3})=-20$ kJ.

Solution: In an isochoric process the volume is constant, so the work done is zero, $W_{isochoric}=0$. Work done on the gas equals the area under the curve, and for compression to half the volume $W_{adiabatic}>W_{isothermal}>W_{isobaric}$. Hence the adiabatic process results in the maximum work done on the gas.

Solution: For isothermal expansion $pV=p'\times2V\Rightarrow p'=\dfrac{p}{2}$. For adiabatic expansion $pV^{\gamma}=$ constant, so $p'(2V)^{5/3}=p''(16V)^{5/3}$, giving $p''=\dfrac{p}{2}\left(\dfrac{2V}{16V}\right)^{5/3}=\dfrac{p}{2}\left(\dfrac{1}{8}\right)^{5/3}=\dfrac{p}{2}\left(\dfrac{1}{32}\right)=\dfrac{p}{64}$.

Solution: Work done in a cyclic process = area bounded by the closed configuration. Here the closed path consists of two equal triangles of opposite sense, so the net area $=\dfrac{1}{2}\times V_0\times p_0-\dfrac{1}{2}\times V_0\times p_0=0$ (zero).

Solution: Net work done = area enclosed by the $pV$ curve = area of $\triangle ABC=\dfrac{1}{2}\times5\times10^{-3}\times4\times10^{5}=10^{3}$ J $=1000$ J.

Solution: For a cyclic process $\Delta U=0\Rightarrow Q=W$. Work $W=-$(area enclosed)$=-AB\times AD=-(2p-p)(3V-V)=-p\times2V=-2pV$. Hence heat rejected $=2pV$.

Solution: The gas first expands isothermally from $V$ to $3V$ (a rectangular hyperbola in the p-V plane), then its volume is reduced from $3V$ to $V$ at constant pressure (a horizontal line). Diagram (d) represents this correctly.

Solution: In an isochoric process the volume (not the pressure) remains constant, so statement (b) is not true. The other statements correctly describe adiabatic and isothermal processes.

Solution: Transfer of heat by radiation and by conduction, and electrical (resistive) heating of a nichrome wire, are all irreversible. Only isothermal compression (as in a Carnot cycle) is a reversible process.

Solution: For an adiabatic process $Q=0$, so $W=-\Delta U=\dfrac{nR}{\gamma-1}(T_i-T_f)$. With $n=1$, $W=6R$, $\gamma=\dfrac{5}{3}$: $6R=\dfrac{R}{\frac{5}{3}-1}(T-T_f)=\dfrac{3R}{2}(T-T_f)\Rightarrow T-T_f=4\Rightarrow T_f=(T-4)$ K.

Solution: $\Delta U=\dfrac{1}{\gamma-1}(p_2V_2-p_1V_1)$. At constant pressure with $V_1=V,\ V_2=2V$: $\Delta U=\dfrac{1}{\gamma-1}(p\times2V-p\times V)=\dfrac{pV}{\gamma-1}$.

Solution: Work done equals the area under the p-V curve. For the same volume change the isobaric path encloses the largest area with the volume axis (its slope is least steep), so $W$ is greatest for the isobaric expansion.

Solution: For an adiabatic process $pV^{\gamma}=k$. Using $V=\dfrac{RT}{p}$, $p\left(\dfrac{RT}{p}\right)^{\gamma}=k\Rightarrow p^{1-\gamma}T^{\gamma}=$ constant.

Solution: In an isothermal process the temperature is constant ($\Delta T=0$), so the internal energy does not change. Hence the increase in internal energy is zero for both gases.