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en/5.6.15.md
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| + | ### Statement | ||
| + | |||
| + | $5.6.15.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | Assuming we're working with ideal gases\ | ||
| + | We have:\ | ||
| + | $V_1=2L=2 \cdot 10^{-3} m^3$ | ||
| + | $P_1=0.8MPa$=0.8 \cdot 10^6 Pa$\ | ||
| + | $V_2=10L=10 \cdot 10^{-3} m^3$\ | ||
| + | for an isothermical process, the work done is\ | ||
| + | $W=nRT\ln(\frac{V_2}{V_1})$\ | ||
| + | but $P_1V_1=nRT$ , then\ | ||
| + | $W=P_1V_1\ln(\frac{V_2}{V_1})$\ | ||
| + | and calculating we get\ | ||
| + | $W \approx 2575J$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $5.6.15.$ [Insert the problem statement] | |||
| ### Solution | |||
| Assuming we're working with ideal gases\ | |||
| We have:\ | |||
| $V_1=2L=2 \cdot 10^{-3} m^3$ | |||
| $P_1=0.8MPa$=0.8 \cdot 10^6 Pa$\ | |||
| $V_2=10L=10 \cdot 10^{-3} m^3$\ | |||
| for an isothermical process, the work done is\ | |||
| $W=nRT\ln(\frac{V_2}{V_1})$\ | |||
| but $P_1V_1=nRT$ , then\ | |||
| $W=P_1V_1\ln(\frac{V_2}{V_1})$\ | |||
| and calculating we get\ | |||
| $W \approx 2575J$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||