Question

An ideal gas is expanded so that amount of heat given is equal to the decrease in internal energy. The gas undergoes the process TV^1/5 = constant. The adiabatic compressibility of gas when pressure is P, is –

• A. $\frac{7}{5P}$
• B. $\frac{5}{7P}$
• C. $\frac{2}{5P}$
• D. $\frac{7}{3P}$

Answer 1

Answer: (B)

$\frac{5}{7P}$

Answer 2

dQ = – dU

C = –C_V = $\frac{–R}{\gamma –1}$ = $\frac{+ R}{\gamma –1}$ + $\frac{dV}{dT}$

–\frac{P}{n}\frac{dV}{dT} = \frac{2R}{\gamma –1} \end{matrix}$$ T⁵V = const. V = $\frac{const.}{T^{5}}$ $\frac{dV}{dT}$ = – 5$\frac{const}{T^{6}}$ PV = nRT P/n = RT/V \+ $\frac{RT}{const.}T^{5}$ × $\left( –5\frac{const}{T^{6}} \right)$= $\frac{2R}{\gamma –1}$ $\frac{5}{2}$= $\frac{1}{\gamma –1}$ Ž g – 1 = 2/5 g = 7/5 adiabatic compressibility b = $\frac{1}{\gamma P}$ = $\frac{5}{7P}$