Question

What is minimum pressure (in \[{\text{ }}kPa\]) a given volume of an ideal gas ${V_2}$ and ${V_1}$ originally at \[400K\] and \[100kPa\] pressure can be compressed irreversibly adiabatically in order raise its temperature to \[600K\]:
\[\begin{array}{*{20}{l}}
  {\left( A \right){\text{ }}362.5{\text{ }}kPa} \\
  {\left( B \right){\text{ }}275{\text{ }}kPa} \\
  {\left( C \right){\text{ }}437.5{\text{ }}kPa} \\
  {\left( D \right){\text{ }}550{\text{ }}kPa}
\end{array}\]

Answer 1

Hint: To get the lowest pressure, we will first apply the formula, and in the adiabatic irreversible process, we will utilise the first rule of thermodynamics, which states that the change in the internal energy of a closed system equals the total of the heat contributed to the system and the work done. Then we use the ideal gas law is \[PV = nRT\] only valid in equilibrium and then we find \[{C_V}\] by the heat capacity relation after then putting the values we find the correct answer.
Formula used:
\[ \Rightarrow n{C_V}\Delta T = - {P_{ext}}({V_2} - {V_1})\]
Here \[{C_V}\] is molar heat capacity when heat is constant
\[\Delta T\]is a change in Temperature.
\[{P_{ext}}\] is external pressure.
\[({V_2} - {V_1})\] is change in volume in gas

Complete step-by-step solution:
Adiabatic process is a type of thermodynamic process which occurs without transferring heat or mass between the system and its surroundings. An adiabatic process, unlike an isothermal process, only sends energy to the environment as work.
In the adiabatic is in irreversible process in which compression is there
So, in this case change in internal energy is given as:
$ \Rightarrow \Delta U = {\text{work done (}}W{\text{)}}$
Work defined as the amount of energy necessary to move something against the pull of gravity. Work and other forms of energy transfer, such as heat, can modify the energy of a system.
Therefore, using the formula
\[ \Rightarrow n{C_V}({T_2} - {T_2}) = - {P_{ext}}({V_2} - {V_1})\]
Here \[{P_{ext}} = {P_2}\] from the question
\[ \Rightarrow n{C_V}({T_2} - {T_2}) = - {P_2}({V_2} - {V_1})\] ------------ eq(i)
And from the ideal gas equation\[PV = nRT\] we write:
\[ \Rightarrow {V_2} = \dfrac{{nR{T_2}}}{{{P_2}}}\]
\[ \Rightarrow {V_1} = \dfrac{{nR{T_1}}}{{{P_1}}}\]
We put the values of ${V_2}$ and ${V_1}$ in eq (i) and we get
\[ \Rightarrow n{C_V}({T_2} - {T_2}) = - {P_2}(\dfrac{{nR{T_2}}}{{{P_2}}} - \dfrac{{nR{T_1}}}{{{P_1}}})\]---(ii)
First, we have to find \[{C_V}\] so from the heat capacity relation
\[ \Rightarrow {C_p} - {C_v} = R\]Here R is universal gas constant
\[ \Rightarrow {C_v} = 5R/2\] Now we put \[{C_V}\] in equation(ii)
\[ \Rightarrow \dfrac{5}{{2}}nR({T_2} - {T_2}) = - {P_2}(\dfrac{{nR{T_2}}}{{{P_2}}} - \dfrac{{nR{T_1}}}{{{P_1}}})\]
\[ \Rightarrow \dfrac{5}{2}({T_2} - {T_2}) = - {P_2}(\dfrac{{{T_2}}}{{{P_2}}} - \dfrac{{{T_1}}}{{{P_1}}})\]
Now putting the values given in the question
\[ \Rightarrow \dfrac{5}{2}(600 - 400) = - {P_2}(\dfrac{{600}}{{{P_2}}} - \dfrac{{400}}{{100}})\]
\[ \Rightarrow 500 = - 600 + 4P\]
\[ \Rightarrow {P_2} = {\text{275 }}kPa\]
Hence, the correct answer is (B).

Note:Even though the ideal gas equation isn't accurate in many circumstances, one can draw something from it. That pressure is directly related to pressure when the volume is kept constant. So, to achieve low pressure/temperature, one of them must be reduced.
When an ideal gas is compressed adiabatically $Q = 0$ , It puts to work, and its temperature rises; in an adiabatic expansion, the gas puts to work, and its temperature falls.