Q amount of heat is given to 0.5 text{ mole} | vedclass

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(B) For a polytropic process $TV^n = 	ext{constant}$, the molar heat capacity is given by $C = C_V + \frac{R}{1-n} = \frac{3R}{2} + \frac{R}{1-n}$.Gi

Vedclass · Accepted Answer

$\frac{-2}{33}$

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(B) For a polytropic process $TV^n = 	ext{constant}$, the molar heat capacity is given by $C = C_V + \frac{R}{1-n} = \frac{3R}{2} + \frac{R}{1-n}$.Given $\Delta Q = n_{m} C \Delta T$, where $n_{m} = 0.5 	ext{ mole}$ and $\Delta T = 227^{\circ}	ext{C} - 27^{\circ}	ext{C} = 200 	ext{ K}$.Total heat $\Delta Q = 15 	imes 10^3 	ext{ J}$.$15000 = 0.5 	imes \left( \frac{3R}{2} + \frac{R}{1-n} ight) 	imes 200$.$15000 = 100 	imes \left( \frac{3R}{2} + \frac{R}{1-n} ight) \implies 150 = \frac{3R}{2} + \frac{R}{1-n}$.Taking $R = \frac{25}{3} 	ext{ J/mol K}$, we get $150 = \frac{3}{2} 	imes \frac{25}{3} + \frac{25/3}{1-n} = 12.5 + \frac{25}{3(1-n)}$.$137.5 = \frac{25}{3(1-n)} \implies 5.5 = \frac{1}{3(1-n)}$.$16.5(1-n) = 1 \implies 1 - n = \frac{1}{16.5} = \frac{2}{33}$.$n = 1 - \frac{2}{33} = \frac{31}{33}$.Wait, re-evaluating the process equation $TV^n = 	ext{constant}$. The standard polytropic form is $PV^x = 	ext{constant}$, which implies $TV^{x-1} = 	ext{constant}$. Thus $n = x-1$. Using $C = \frac{R}{\gamma-1} + \frac{R}{1-x} = \frac{3R}{2} + \frac{R}{1-x}$.From $150 = 12.5 + \frac{25}{3(1-x)}$, we get $1-x = \frac{25}{3 	imes 137.5} = \frac{25}{412.5} = \frac{2}{33}$.$x = 1 - \frac{2}{33} = \frac{31}{33}$. Since $n = x-1$, $n = \frac{31}{33} - 1 = \frac{-2}{33}$.

$Q$ amount of heat is given to $0.5 \text{ mole}$ of an ideal mono-atomic gas by a process $TV^n = \text{constant}$. The following graph shows the variation of temperature with $Q$. Find the value of $n$.

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