Question

Two moles of an ideal gas $ ({C_{v.m}} = 3/2{\text{R}})\; $ is subjected to a following change of state:
 $ {\text{A}}\left( {500{\text{K}},5.0{\text{bar}}} \right)\xrightarrow[{{\text{isothermal expansion}}}]{{{\text{reversible}}}}{\text{B}}\xrightarrow[{{\text{cooling}}}]{{{\text{isochoric}}}}{\text{C}}\left( {250{\text{K}},1{\text{bar}}} \right)\xrightarrow[{{\text{adiabatic compression}}}]{{{\text{single stage}}}}{\text{D}}\left( {3{\text{bar}}} \right) $
The correct statement is are:
A.The pressure at B is $ {\text{2}}{\text{.0bar}} $
B.The temperature at D is $ {\text{450K}} $
C. $ \Delta {{\text{H}}_{{\text{CD}}}} = 1000{\text{R}} $
D. $ \Delta {\text{U}} = 375{\text{R}} $

Answer 1

Hint: To answer this question, you should recall the concept of adiabatic compression, isothermal expansion and isothermal cooling. We shall substitute the appropriate values in the formula to calculate the answer.

The formula used:
1) $ {{\text{C}}_{\text{p}}} - {{\text{C}}_{\text{v}}} = {\text{R; }} $ where $ {{\text{C}}_{\text{p}}} $ is the molar specific heat capacity of an ideal gas at constant pressure, $ {{\text{C}}_{\text{v}}} $ is its molar specific heat at constant volume and R is the gas constant.
2) $ {\text{PV}} = {\text{nRT}} $ where P is pressure, V is volume, R is the universal gas constant, n is no. of moles and T is temperature

Complete step-by-step answer:Let initial volume be $ {\text{V}} $ . So, moving from C to D
 $ {{\text{C}}_{{\text{v}}}} = 3/2{\text{R}} $
and putting it into the formula we have:
  $ \Rightarrow {{\text{C}}_{\text{p}}} = 5/2{\text{R}} $ $ \& {\text{ }}\gamma = {\text{Cp}}{\text{/Cv}} $
 $ {\text{ }}\; \Rightarrow {\text{ }}\;\gamma = 5/3 $
The transition from A to B is isothermal that means the temperature at both A and B is the same = $ 500{\text{K}} $ .
Also, the transition from B to C is isochoric that is volume constant:
 $ \dfrac{{{{\text{P}}_{\text{B}}}}}{{{{\text{T}}_{\text{B}}}}} = \dfrac{{{{\text{P}}_{\text{c}}}}}{{{{\text{T}}_{\text{c}}}}} $
Putting the values as given in the question:
 $ \dfrac{{{{\text{P}}_{\text{B}}}}}{{500}} = \dfrac{1}{{250}} $ .
 Cross multiplying and solving:
 $ \Rightarrow {{\text{P}}_{\text{B}}} = 2{\text{ bar}} $
Also for process CD which is adiabatic, that means heat exchange is zero
Hence, change in internal energy = Work done
 $ {\text{n}}{{\text{C}}_{\text{V}}}{{\Delta T}} = - {{\text{p}}_{{\text{ext}}}}{{\Delta V}} $ .
 $ \Rightarrow {\text{n}}{{\text{C}}_{\text{V}}}\left( {{{\text{T}}_{\text{D}}} - {{\text{T}}_{\text{C}}}} \right) = - {{\text{p}}_{{\text{ext}}}}\left[ {\dfrac{{{\text{nR}}{{\text{T}}_{\text{D}}}}}{{{{\text{p}}_{\text{d}}}}} - \dfrac{{{\text{nR}}{{\text{T}}_c}}}{{{{\text{p}}_{\text{c}}}}}} \right] $ .
Using the ideal gas equation to substitute the work done given in the equation: Putting the values of the variables in the above equation
 $ {\text{d}}\dfrac{3}{2}\left( {{{\text{T}}_{\text{B}}} - {{\text{T}}_{\text{C}}}} \right) = - 3\left[ {\dfrac{{{{\text{T}}_{\text{D}}}}}{3} - \dfrac{{{{\text{T}}_{\text{c}}}}}{1}} \right] $ .
After solving:
 $ {{\text{T}}_{\text{D}}} = 450{\text{K}} $ .
Hence, option A, B, C is correct
Also, $ \Delta {{\text{H}}_{{\text{CD}}}} = {\text{n}}{{\text{C}}_{\text{p}}}{{\Delta T = }}2 \times \dfrac{5}{2}{\text{R}} \times 200 = 1000{\text{R}} $

Note: The student should not confuse between other types of processes:
Isothermal process: When the system undergoes change from one state to the other, but its temperature remains constant, the system is said to have undergone an isothermal process
Adiabatic process: The process, during which the heat content of the system or a certain quantity of the matter remains constant, is called an adiabatic process.
Isochoric process: The process, during which the volume of the system remains constant, is called an isochoric process. Heating of gas in a closed cylinder is an example of the isochoric process.
Isobaric process: The process during which the pressure of the system remains constant is called an isobaric process.