A) 90.5 J
B) 60.7 J
C) 48 J
D) 100.8 J
Correct Answer: A
Solution :
| [a] \[{{\text{V}}_{\text{1}}}\text{=1}\,\,\text{litre,}\,\,{{\text{P}}_{\text{1}}}\text{=1}\,\,\text{atm}\] |
| \[{{\text{V}}_{2}}\text{=3}\,\,\text{litre,}\,\,\gamma \text{=1}\text{.40}\], |
| \[{{P}_{2}}V_{2}^{\gamma }={{P}_{1}}V_{1}^{\gamma }\] |
| \[\Rightarrow {{P}_{2}}={{P}_{1}}\times {{\left( \frac{1}{3} \right)}^{1.4}}=\frac{1}{4.6555}atm\] |
| \[\therefore w=\frac{{{P}_{1}}{{V}_{1}}-{{P}_{2}}{{V}_{2}}}{\gamma -1}\] |
| \[=\frac{\left( 1\times 1-\frac{1}{4.6555}\times 3 \right)1.01325\times {{10}^{5}}\times {{10}^{-3}}}{0.4}\] |
| \[=90.1\,\,J\] |
| Closest value of \[W=90.5\,\,J\] |
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