A) 20atm
B) 66.28 atm
C) 30 atm
D) 150 atm
Correct Answer: B
Solution :
\[{{P}_{2}}{{V}_{2}}^{\gamma }={{P}_{1}}{{V}_{1}}^{\gamma }\] \[{{P}_{2}}={{P}_{1}}{{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma }}=1{{\left( \frac{{{V}_{1}}}{{{V}_{1}}/20} \right)}^{1.4}}\] \[\text{=(20}{{\text{)}}^{\text{1}\text{.4}}}\text{=66}\text{.28}\,\text{atm}\]
You need to login to perform this action.
You will be redirected in
3 sec
