A) \[475\,{}^\circ C\]
B) \[402\,{}^\circ C\]
C) \[275\,{}^\circ C\]
D) \[375\,{}^\circ C\]
Correct Answer: D
Solution :
| In adiabatic process |
| \[P{{V}^{\gamma }}=\]constant ...(i) |
| Ideal gas equation is, |
| \[PV=RT\] (for one mole) |
| or \[P=\frac{RT}{V}\] ...(ii) |
| (\[R=\] gas constant) |
| From Eqs. (i) and (ii), we have |
| \[\left( \frac{RT}{V} \right){{V}^{\gamma }}=\text{constant}\] |
| \[T\,\,{{V}^{\gamma -1}}=\] constant |
| so \[{{T}_{1}}V_{1}^{\gamma -1}={{T}_{2}}V_{2}^{\gamma -1}\] |
| or \[\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}}\] ...(iv) |
| Given, \[{{T}_{1}}=\text{ }27{}^\circ C=27+273=300\text{ }K\] |
| \[\frac{{{V}_{2}}}{{{V}_{1}}}=\frac{8}{27},\,\gamma =\frac{5}{3}\] |
| Substituting in Eq. (i), we get |
| \[\frac{{{T}_{2}}}{300}={{\left( \frac{27}{8} \right)}^{5/3-1}}\] |
| or \[\frac{{{T}_{2}}}{300}={{\left[ {{\left( \frac{3}{2} \right)}^{3}} \right]}^{2/3}}\] |
| or \[\frac{{{T}_{2}}}{300}={{\left( \frac{3}{2} \right)}^{2}}=\frac{9}{4}\] |
| \[\therefore \] \[{{T}_{2}}=\frac{9}{4}\times 300=675\,K={{402}^{o}}C\] |
| Thus, rise in temperature |
| \[={{T}_{2}}-{{T}_{1}}=402-27={{375}^{o}}C\] |
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