Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Heat and Thermodynamics question

2020 · 4 Sep · Shift 2 · Q42
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Heat and Thermodynamics
  5. /2020 · 4 Sep · Shift 2 · Q42

Heat and Thermodynamics question

2020 · 4 Sep · Shift 2 · Q42

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1
The change in the magnitude of the volume of an ideal gas when a small additional pressure Δ\DeltaΔ P is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity Δ\DeltaΔ T at constant pressure. The initial temperature and pressure of the gas were 300 K and 2 atm. respectively. If |Δ\DeltaΔ T| = C|Δ\DeltaΔ P| then value of C in (K/atm.) is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 150

  1. For an ideal gas, PV=nRTPV=nRTPV=nRT

  2. We compare two small changes in volume from the same initial state:

  • due to a small pressure change ΔP\Delta PΔP at constant temperature,
  • due to a small temperature change ΔT\Delta TΔT at constant pressure.

Since the magnitudes of volume changes are equal, we use differentials.

  1. At constant temperature: PV=constantPV=\text{constant}PV=constant So, V=nRTPV=\frac{nRT}{P}V=PnRT​ Differentiating w.r.t. PPP at constant TTT, dV=−nRTP2dP=−VPdPdV=-\frac{nRT}{P^2}dP=-\frac{V}{P}dPdV=−P2nRT​dP=−PV​dP Hence magnitude of volume change is ∣ΔV∣P=VP∣ΔP∣|\Delta V|_P=\frac{V}{P}|\Delta P|∣ΔV∣P​=PV​∣ΔP∣

  2. At constant pressure: V=nRTPV=\frac{nRT}{P}V=PnRT​ Differentiating w.r.t. TTT at constant PPP, dV=nRPdT=VTdTdV=\frac{nR}{P}dT=\frac{V}{T}dTdV=PnR​dT=TV​dT Hence magnitude of volume change is ∣ΔV∣T=VT∣ΔT∣|\Delta V|_T=\frac{V}{T}|\Delta T|∣ΔV∣T​=TV​∣ΔT∣

  3. Given these magnitudes are equal: VP∣ΔP∣=VT∣ΔT∣\frac{V}{P}|\Delta P|=\frac{V}{T}|\Delta T|PV​∣ΔP∣=TV​∣ΔT∣ Cancel VVV: ∣ΔT∣∣ΔP∣=TP\frac{|\Delta T|}{|\Delta P|}=\frac{T}{P}∣ΔP∣∣ΔT∣​=PT​ Thus, C=TPC=\frac{T}{P}C=PT​

  4. Substitute the initial values: T=300 K,P=2 atmT=300\,\text{K},\quad P=2\,\text{atm}T=300K,P=2atm So, C=3002=150 K/atmC=\frac{300}{2}=150\,\text{K/atm}C=2300​=150K/atm

Therefore, the required integer is 150\boxed{150}150​

PreviousNext

More from Heat and Thermodynamics

  • Match the thermodynamic processes taking place in a system with the correct conditions. In the table : Δ Q is the heat supplied, Δ W is the work done and Δ U is change in internal energy of the system. Includes table2020 · MCQ
  • A bullet of mass 5 g, travelling with a speed of 210 m/s, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature…2020 · MCQ
  • Number of molecules in a volume of 4 cm3 of a perfect monoatomic gas at some temperature T and at a pressure of 2 cm of mercury is close to? (Given, mean kinetic energy of a molecule (at T) is 4 × 10–14 erg, g = 980 cm/s2, density…2020 · MCQ
  • Three different processes that can occur in an ideal monoatomic gas are shown in the P vs V diagram. The paths are labelled as A → B, A → C and A → D. The change in internal energies during these process are taken as EAB, EAC… Includes diagram2020 · MCQ
  • Nitrogen gas is at 300oC temperature. The temperature (in K) at which the rms speed of a H2 molecule would be equal to the rms speed of a nitrogen molecule, is ​. (Molar mass of N2 gas 28 g).2020 · Numerical
  • In an adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The final pressure of the gas is found to be n times the initial pressure. The value of n is :2020 · MCQ
  • Two different wires having lengths L1 and L2, and respective temperature coefficient of linear expansion α 1 and α 2, are joined end-to-end. Then the effective temperature coefficient of linear expansion is :2020 · MCQ
  • Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom.The gas is maintained at a temperature of T. The total internal energy, U of a mole of this gas, and the value of γ(=Cv​Cp​​)…2020 · MCQ