Zeroth and First Law of Thermodynamics — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Zeroth and First Law of Thermodynamics MCQs with step-by-step solutions (9 questions). Part of Thermodynamics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Zeroth and First Law of Thermodynamics · medium · numerical
$1$ g of water, of volume $1\,\text{cm}^3$ at $100^\circ$C is converted into steam at same temperature under normal atmospheric pressure $=(\simeq1\times10^5$ Pa$)$. The volume of steam formed equals $1671\,\text{cm}^3$. If the specific latent heat of vaporisation of water is $2256$ J/g, the change in internal energy is
A. $2423$ J
B. $2089$ J ✓ Correct
C. $167$ J
D. $2256$ J
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.
Q2 — Zeroth and First Law of Thermodynamics · medium · numerical
A sample of $0.1$ g of water at $100^\circ$C and normal pressure $(1.013\times10^5\,\text{Nm}^{-2})$ 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
A. $42.2$ J
B. $208.7$ J ✓ Correct
C. $104.3$ J
D. $84.5$ 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.
Q3 — Zeroth and First Law of Thermodynamics · medium · numerical
During an isothermal expansion, a confined ideal gas does $-150$ J of work against its surroundings. This implies that
A. $300$ J of heat has been added to the gas
B. no heat is transferred because the process is isothermal
C. $150$ J of heat has been added to the gas ✓ Correct
D. $150$ J of heat has been removed from the gas
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.
Q4 — Zeroth and First Law of Thermodynamics · medium · theory
If $\Delta U$ and $\Delta W$ represent the increase in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true?
A. $\Delta U=-\Delta W$, in an adiabatic process ✓ Correct
B. $\Delta U=\Delta W$, in an isothermal process
C. $\Delta U=\Delta W$, in an adiabatic process
D. $\Delta U=-\Delta W$, in an isothermal process
Solution: From the first law of thermodynamics $\Delta Q=\Delta U+\Delta W$. For an adiabatic process $\Delta Q=0$, therefore $\Delta U=-\Delta W$.
Q5 — Zeroth and First Law of Thermodynamics · medium · numerical
The internal energy change in a system that has absorbed $2$ kcal of heat and done $500$ J of work is
A. $8900$ J
B. $6400$ J
C. $5400$ J
D. $7900$ J ✓ Correct
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.
Q6 — Zeroth and First Law of Thermodynamics · medium · theory
If $Q,E$ and $W$ denote respectively the heat added, change in internal energy and the work done in a closed cycle process, then
A. $W=0$
B. $Q=W=0$
C. $E=0$ ✓ Correct
D. $Q=0$
Solution: For a cyclic process, the internal energy returns to its initial value, so $\Delta U=0$, i.e. $E=0$.
Q7 — Zeroth and First Law of Thermodynamics · medium · theory
We consider a thermodynamic system. If $\Delta U$ represents the increase in its internal energy and $W$ the work done by the system, which of the following statements is true?
A. $\Delta U=-W$ in an adiabatic process ✓ Correct
B. $\Delta U=W$ in an isothermal process
C. $\Delta U=-W$ in an isothermal process
D. $\Delta U=W$ in an adiabatic process
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 $\Delta U=-W$.
Q8 — Zeroth and First Law of Thermodynamics · medium · numerical
$110$ J of heat is added to a gaseous system, whose internal energy is $40$ J, then the amount of external work done is
A. $150$ J
B. $70$ J ✓ Correct
C. $110$ J
D. $40$ J
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.
Q9 — Zeroth and First Law of Thermodynamics · medium · theory
First law of thermodynamics is a consequence of conservation of
A. work
B. energy ✓ Correct
C. heat
D. All of these
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.