A) 28 atm
B) 68.7 atm
C) 256 atm
D) 8 atm
Correct Answer: C
Solution :
\[{{T}_{i}}=273+27=300\,K\] \[{{T}_{2}}=273+927=1200\,K\] Gas equation for adiabatic process \[p{{V}^{\gamma }}\] constant \[p{{\left( \frac{T}{p} \right)}^{\gamma }}=\text{constant}\]\[(\because pV=RT)\] \[\therefore \] \[\frac{{{p}_{2}}}{{{p}_{1}}}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{\frac{\gamma }{\gamma -1}}}\]or\[{{p}_{2}}={{p}_{1}}{{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{\frac{\gamma }{\gamma -1}}}\] \[{{p}_{2}}=2{{\left( \frac{1200}{300} \right)}^{\frac{1.4}{1.4-1}}}\] \[{{p}_{2}}=256\,\text{atm}\]You need to login to perform this action.
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