JEE Challenger
More from Thermodynamics
ChemMedAdv 2024 • Paper 1

Change in Enthalpy for Ideal Monoatomic Gas Expansion in VT Diagram

Consider the following volume–temperature (V–T) diagram for the expansion of 5 moles of an ideal monoatomic gas.

Considering only P-V work is involved, the total change in enthalpy (in Joule) for the transformation of state in the sequence X→Y→Z\mathbf{X}\rightarrow\mathbf{Y}\rightarrow\mathbf{Z} is ______.

[Use the given data: Molar heat capacity of the gas for the given temperature range, Cv, m=12 J K−1 mol−1C_{\text{v, m}} = 12\text{ J K}^{-1}\text{ mol}^{-1} and gas constant, R=8.3 J K−1 mol−1R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}]

Question Diagram 1
Official Numerical Answer8120

Topics & Concepts

ThermodynamicsEnthalpyFirst Law of Thermodynamics

Step-by-Step Solution

To find the total change in enthalpy (ΔH\Delta H) for the transformation of state along the sequence X→Y→Z\mathbf{X} \rightarrow \mathbf{Y} \rightarrow \mathbf{Z}, we use the fundamental thermodynamic principle that enthalpy (HH) is a state function for an ideal gas and depends solely on the initial and final temperatures.

1. Identify Given Data

  • Number of moles of gas, n=5 moln = 5\text{ mol}
  • Molar heat capacity at constant volume, Cv,m=12 J K−1 mol−1C_{v, m} = 12\text{ J K}^{-1}\text{ mol}^{-1}
  • Universal gas constant, R=8.3 J K−1 mol−1R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}

2. Molar Heat Capacity at Constant Pressure (Cp,mC_{p, m})

For an ideal gas, the relationship between Cp,mC_{p, m} and Cv,mC_{v, m} is given by Mayer's relation: Cp,m=Cv,m+RC_{p, m} = C_{v, m} + R

Substituting the given values: Cp,m=12+8.3=20.3 J K−1 mol−1C_{p, m} = 12 + 8.3 = 20.3\text{ J K}^{-1}\text{ mol}^{-1}

3. Temperature of States from the V–TV\text{--}T Diagram

  • State X\mathbf{X}: Temperature TX=335 KT_X = 335\text{ K}, Volume VX=10 LV_X = 10\text{ L}
  • State Y\mathbf{Y}: Temperature TY=335 KT_Y = 335\text{ K}, Volume VY=20 LV_Y = 20\text{ L}
  • State Z\mathbf{Z}: Temperature TZ=415 KT_Z = 415\text{ K}, Volume VZ=20 LV_Z = 20\text{ L}

4. Calculation of Enthalpy Change (ΔH\Delta H)

Since X→Y\mathbf{X} \rightarrow \mathbf{Y} is an isothermal process (TX=TY=335 KT_X = T_Y = 335\text{ K}): ΔHX→Y=nCp,m(TY−TX)=nCp,m(335−335)=0 J\Delta H_{\mathbf{X} \rightarrow \mathbf{Y}} = n C_{p, m} (T_Y - T_X) = n C_{p, m} (335 - 335) = 0\text{ J}

For the isochoric process Y→Z\mathbf{Y} \rightarrow \mathbf{Z}: ΔHY→Z=nCp,m(TZ−TY)\Delta H_{\mathbf{Y} \rightarrow \mathbf{Z}} = n C_{p, m} (T_Z - T_Y) ΔHY→Z=5×20.3×(415−335)\Delta H_{\mathbf{Y} \rightarrow \mathbf{Z}} = 5 \times 20.3 \times (415 - 335) ΔHY→Z=5×20.3×80=8120 J\Delta H_{\mathbf{Y} \rightarrow \mathbf{Z}} = 5 \times 20.3 \times 80 = 8120\text{ J}

5. Total Change in Enthalpy

ΔHtotal=ΔHX→Y+ΔHY→Z=0+8120=8120 J\Delta H_{\text{total}} = \Delta H_{\mathbf{X} \rightarrow \mathbf{Y}} + \Delta H_{\mathbf{Y} \rightarrow \mathbf{Z}} = 0 + 8120 = 8120\text{ J}

Alternatively, directly using the state function property from X\mathbf{X} to Z\mathbf{Z}: ΔHtotal=nCp,m(TZ−TX)=5×20.3×(415−335)=8120 J\Delta H_{\text{total}} = n C_{p, m} (T_Z - T_X) = 5 \times 20.3 \times (415 - 335) = 8120\text{ J}

8120

Change in Enthalpy for Ideal Monoatomic Gas Expansion in VT Diagram | Chemistry PYQ Solution - JEE Challenger