A) \[{{475}^{\text{o}}}C\]
B) \[{{402}^{\text{o}}}C\]
C) \[{{275}^{\text{o}}}C\]
D) \[{{375}^{\text{o}}}C\]
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, \[{{T}_{1}}={{27}^{o}}C\,=27+273\,=300\,K\] \[\frac{{{V}_{2}}}{{{V}_{1}}}=\frac{8}{27},\,\gamma =\frac{5}{3}\] 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 \[={{T}_{2}}-{{T}_{1}}=402-27={{375}^{o}}C\]
You need to login to perform this action.
You will be redirected in
3 sec
