A) - 46 kJ, 46 kJ
B) 36 kJ, - 36 kJ
C) 46 kJ, - 46 kJ
D) - 36 kJ, 36 kJ
Correct Answer: A
Solution :
[a]Let \[{{p}_{1}}=1\] atm, \[n=5\,mol,\text{ }293K\] |
\[{{V}_{2}}=\frac{{{V}_{1}}}{10}\] |
Using \[{{T}_{1}}V_{1}^{\gamma -1}={{T}_{2}}V_{2}^{\gamma -1}\] |
\[\Rightarrow \] \[{{T}_{2}}={{T}_{1}}{{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}}\] |
\[=293{{(10)}^{0.4}}=736K\] |
Now, work done \[=\frac{nR({{T}_{1}}-{{T}_{2}})}{\gamma -1}\] |
\[=\frac{5\times 8.3\times (293-736)}{0.4}=-46kJ\] |
and \[\Delta U=\Delta Q-W=0-W=46kJ\] |
You need to login to perform this action.
You will be redirected in
3 sec