A) \[{{V}^{a-b}}.{{T}^{b}}.{{e}^{cT}}=\text{constant}\]
B) \[{{T}^{a-b}}.{{V}^{b}}.{{e}^{cT}}=\text{constant}\]
C) \[{{T}^{a-b}}.{{V}^{a}}.{{e}^{cT}}=\text{constant}\]
D) \[{{V}^{a-b}}.{{T}^{a}}.{{e}^{cT}}=\text{constant}\]
Correct Answer: A
Solution :
| For adiabatic process, \[\Delta Q=\Delta U=\Delta W=0\] |
| \[\therefore \,\,\,\,{{C}_{V}}dT+pdV=0\]\[\Rightarrow \,\,{{C}_{V}}dT+\frac{RT}{V}.dV=0\]\[\Rightarrow \,\,{{C}_{V}}dT+\left( {{C}_{p}}-{{C}_{v}} \right)T.\frac{dV}{V}=0\] |
| \[\left( b+cT \right)dT+\left( a-b \right)T.\frac{dV}{V}=0\]\[=b\frac{dT}{T}+cdT+\left( a-b \right)\frac{dV}{V}=0\] |
| Integrating above equation, we have |
| \[b\ln T+cT+\left( a-b \right)InV=a\] constant So,\[ln\,\,{{T}^{b}}+cT+\ln \,{{V}^{a-b}}=a\] constant\[\Rightarrow \,\,\,{{V}^{a-b}}.{{T}^{b}}.{{e}^{cT}}=cons\tan t\] |
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