Правка разделов «Statement», «Solution»
en/5.6.17.md
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| @@ -1,16 +1,19 @@ | |||
| ### Statement | |||
| − | $5.6.17.$ | ||
| + | $5.6.17.$ One mole of hydrogen having a temperature of 0 | ||
| + | ◦C is heated at constant pressure. How much heat must be transferred to the gas to double its volume? | ||
| + | What kind of work will be performed by gas? | ||
| ### Solution | |||
| Assuming that hydrogen (H_2) behaves has an ideal gas\ | |||
| We have\ | |||
| $n=1mol$\ | |||
| − | $T=0 | ||
| + | $T=0^\circ=273K$\ | ||
| $P=const$\ | |||
| − | Q=nC_p\Delta T\ | ||
| − | |||
| + | $Q=nC_p\Delta T$\ | ||
| + | we know that\ | ||
| + | $C_p=\frac{(i+2)R}{2}$\ | ||
| where i is the number of degrees of freedom, which is 5 for hydrogen (diatomic gas)\ | |||
| $C_p=\frac{7R}{2}$\ | |||
| so $Q=\frac{7nR\Delta T}{2}$\ | |||
| and with Charles's Law ( $\frac{V}{T}=const$ )\ | |||
| $\frac{V}{T_1}=\frac{2V}{T_2}$\ | |||
| $T_2=2T_1$ , so $\Delta T=T_2-T_1=2T_1-T_1=T_1$\ | |||
| and substituting in the heat\ | |||
| $Q=\frac{7nRT_1}{2}$\ | |||
| $Q\approx 7948.5J$ | |||
| and the work for a constant pressure\ | |||
| $W=P\Delta V=P(2V-V)=PV=nRT_1$\ | |||
| @@ -25,6 +28,3 @@Solution | |||
| $W\approx 2271J$\ | |||
| It's an isobaric expansion work | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,16 +1,19 @@ | |||
| ### Statement | ### Statement | ||
| $5.6.17.$ |
$5.6.17.$ One mole of hydrogen having a temperature of 0 | ||
| ◦C is heated at constant pressure. How much heat must be transferred to the gas to double its volume? | |||
| What kind of work will be performed by gas? | |||
| ### Solution | ### Solution | ||
| Assuming that hydrogen (H_2) behaves has an ideal gas\ | Assuming that hydrogen (H_2) behaves has an ideal gas\ | ||
| We have\ | We have\ | ||
| $n=1mol$\ | $n=1mol$\ | ||
| $T=0 |
$T=0^\circ=273K$\ | ||
| $P=const$\ | $P=const$\ | ||
| Q=nC_p\Delta T\ | $Q=nC_p\Delta T$\ | ||
| we know that\ | |||
| $C_p=\frac{(i+2)R}{2}$\ | |||
| where i is the number of degrees of freedom, which is 5 for hydrogen (diatomic gas)\ | where i is the number of degrees of freedom, which is 5 for hydrogen (diatomic gas)\ | ||
| $C_p=\frac{7R}{2}$\ | $C_p=\frac{7R}{2}$\ | ||
| so $Q=\frac{7nR\Delta T}{2}$\ | so $Q=\frac{7nR\Delta T}{2}$\ | ||
| and with Charles's Law ( $\frac{V}{T}=const$ )\ | and with Charles's Law ( $\frac{V}{T}=const$ )\ | ||
| $\frac{V}{T_1}=\frac{2V}{T_2}$\ | $\frac{V}{T_1}=\frac{2V}{T_2}$\ | ||
| $T_2=2T_1$ , so $\Delta T=T_2-T_1=2T_1-T_1=T_1$\ | $T_2=2T_1$ , so $\Delta T=T_2-T_1=2T_1-T_1=T_1$\ | ||
| and substituting in the heat\ | and substituting in the heat\ | ||
| $Q=\frac{7nRT_1}{2}$\ | $Q=\frac{7nRT_1}{2}$\ | ||
| $Q\approx 7948.5J$ | $Q\approx 7948.5J$ | ||
| and the work for a constant pressure\ | and the work for a constant pressure\ | ||
| $W=P\Delta V=P(2V-V)=PV=nRT_1$\ | $W=P\Delta V=P(2V-V)=PV=nRT_1$\ | ||
| @@ -25,6 +28,3 @@Solution | |||
| $W\approx 2271J$\ | $W\approx 2271J$\ | ||
| It's an isobaric expansion work | It's an isobaric expansion work | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||