Question

In a flexible balloon, 2 moles of \[S{O_2}\] having an initial volume of 1Kl at a temperature of \[{27^\circ }C\] is filled (\[S{O_2}\] is a linear triatomic gas). The gas is first expanded to thrice its initial volume isobarically and then expanded adiabatically so as to attain its initial temperature. Assuming gas is ideal, \[\gamma = \dfrac{4}{3}\] and \[R = \dfrac{{25}}{3}Jmo{l^{ - 1}}{K^{ - 1}}\]. Work done by the gas in the whole process is
A) \[2KJ\]
B) \[3.2KJ\]
C) \[40KJ\]
D) \[15KJ\]

Answer 1

Hint: In the isobaric process, the pressure is kept at constant while transferring heat from one system to another. In the adiabatic process, there is no heat transferred from the system to the surroundings. The heat is kept at constant by changing the temperature. In the adiabatic process, the system is insulated from the surrounding.

Complete step by step answer:
(i) In an isobaric process, the pressure is kept at constant when the volume and temperature of the gas is changed. We know that \[PV = RT\] where R is a gas constant and in isobaric process pressure is also constant. Hence the ratio of volume and temperature \[\dfrac{V}{T}\]=constant.
(ii) Therefore the ratio of volume and temperature before and after the isobaric process is constant. Hence,
\[ \Rightarrow \dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}\]
As the volume is increased thrice, \[{V_2} = 3{V_1}\]. And the initial temperature \[{T_1} = 27^\circ C \Rightarrow (27 + 273)K\] therefore \[{T_1} = 300^\circ C\]
(iii) \[\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}\]
\[ \Rightarrow {T_2} = \dfrac{{{V_2}}}{{{V_1}}}{T_1}\]
\[ \Rightarrow {T_2} = \dfrac{{3{V_1}}}{{{V_1}}}(300)\]
\[ \Rightarrow {T_2} = 900K\]
(iv)The work done at isobaric process,
\[W = nR\Delta T\]
\[ \Rightarrow W = nR({T_2} - {T_1})\]
\[ \Rightarrow W = 2 \times \dfrac{{25}}{3} \times (900 - 300)\]
\[ \Rightarrow W = 9972J\] ----------- (1)
(v) In adiabatic process, the heat is kept at constant by varying the temperature of the system. And \[P{V^\gamma }\]=constant where \[\gamma \] is the adiabatic index which equals the ratio of heat capacity at constant pressure to the heat capacity at the constant volume.
\[P{V^\gamma } = RT\]
And we know that the pressure is inversely proportional to the volume according to Boyle’s law.
\[ \Rightarrow \dfrac{{{V^\gamma }}}{V} = RT\]
\[ \Rightarrow {V^{\gamma - 1}} = RT\]
\[ \Rightarrow \dfrac{{{V^{\gamma - 1}}}}{T}\]=constant. As $R$ is a gas constant.
(vi)Therefore the ratio of volume and temperature before and after adiabatic process is,
\[ \Rightarrow \dfrac{{{V_1}^{\gamma - 1}}}{{{T_1}}} = \dfrac{{{V_2}^{\gamma - 1}}}{{{T_2}}}\]
\[ \Rightarrow {T_2} = {\left( {\dfrac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}}{T_1}\]
\[ \Rightarrow {T_2} = {\left( {\dfrac{1}{3}} \right)^{\dfrac{4}{3} - 1}}(300)\]
\[ \Rightarrow {T_2} = 207.3K\]
(vii) The work done at adiabatic process,
\[ \Rightarrow W = \dfrac{{nR\Delta T}}{{\gamma - 1}}\]
\[ \Rightarrow W = \dfrac{{2 \times \dfrac{{25}}{3}(300 - 207)}}{{\dfrac{4}{3} - 1}}\]
\[ \Rightarrow W = 2 \times 8.31 \times 93 \times 3\]
\[ \Rightarrow W = 4637J\] ------------ (2)
(viii) To find the total work done by the gas in the whole process can be obtained by summing (1) and (2)
\[ \Rightarrow {W_{total}} = 9972 + 4637\]
\[ \Rightarrow {W_{total}} = 14609J\]
\[\therefore W = 14.6KJ \approx 15KJ\]

Hence the 15KJ is the total work done by the gas in the whole process. Therefore the correct option is D.

Note:
(i) Boyle’s law gives the relation between the pressure and volume (P-V) for the ideal gas of an object. It states that the pressure exerted by the object of an ideal gas is inversely proportional to its volume when the temperature and the amount of gas remain unchanged in the system. \[\therefore PV = k\]
(ii) In adiabatic process, \[P{V^\gamma }\] = constant where \[\gamma \] is the adiabatic index which equals to the ratio of heat capacity at constant pressure to the heat capacity at the constant volume \[\left( {\dfrac{{{C_P}}}{{{C_V}}}} \right)\]