Edits to “Statement”, “Solution”, “Answer”
en/5.6.15.md
+5 −6
| @@ -1,21 +1,20 @@ | |||
| ### Statement | |||
| − | $5.6.15.$ | ||
| + | $5.6.15.$ The air, which occupied a volume of 2 liters at a pressure of 0.8 MPa, isothermically expanded to 10 liters. Identify the work done by air. | ||
| ### Solution | |||
| Assuming we're working with ideal gases\ | |||
| We have:\ | |||
| − | $V_1=2L=2 | ||
| − | $P_1=0.8MPa | ||
| + | $V_1=2L=2\cdot 10^{-3} m^3$ | ||
| + | $P_1=0.8MPa=0.8 \cdot 10^6 Pa$\ | ||
| $V_2=10L=10 \cdot 10^{-3} m^3$\ | |||
| − | for an isothermic | ||
| + | for an isothermic 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] | ||
| + | $W \approx 2575J$ | ||
| @@ -1,21 +1,20 @@ | |||
| ### Statement | ### Statement | ||
| $5.6.15.$ |
$5.6.15.$ The air, which occupied a volume of 2 liters at a pressure of 0.8 MPa, isothermically expanded to 10 liters. Identify the work done by air. | ||
| ### Solution | ### Solution | ||
| Assuming we're working with ideal gases\ | Assuming we're working with ideal gases\ | ||
| We have:\ | We have:\ | ||
| $V_1=2L=2 |
$V_1=2L=2\cdot 10^{-3} m^3$ | ||
| $P_1=0.8MPa |
$P_1=0.8MPa=0.8 \cdot 10^6 Pa$\ | ||
| $V_2=10L=10 \cdot 10^{-3} m^3$\ | $V_2=10L=10 \cdot 10^{-3} m^3$\ | ||
| for an isothermic |
for an isothermic process, the work done is\ | ||
| $W=nRT\ln(\frac{V_2}{V_1})$\ | $W=nRT\ln(\frac{V_2}{V_1})$\ | ||
| but $P_1V_1=nRT$ , then\ | but $P_1V_1=nRT$ , then\ | ||
| $W=P_1V_1\ln(\frac{V_2}{V_1})$\ | $W=P_1V_1\ln(\frac{V_2}{V_1})$\ | ||
| and calculating we get\ | and calculating we get\ | ||
| $W \approx 2575J$ | $W \approx 2575J$ | ||
| #### Answer | #### Answer | ||
| $W \approx 2575J$ | |||
| [Insert a concise answer or boxed result] | |||