One mole of an ideal gas left(gamma = frac{5} | vedclass

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(D) Given: $\mu = 1 	ext{ mole}$,$\gamma = \frac{5}{3}$,$T_1 = 127^{\circ} C = 400 \ K$,$V_2 = \frac{8}{27} V_1$.For an adiabatic process,$T_1 V_1^{\

Vedclass · Accepted Answer

$1500$

Vedclass · Answer

(D) Given: $\mu = 1 	ext{ mole}$,$\gamma = \frac{5}{3}$,$T_1 = 127^{\circ} C = 400 \ K$,$V_2 = \frac{8}{27} V_1$.For an adiabatic process,$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.$T_2 = T_1 \left( \frac{V_1}{V_2} ight)^{\gamma-1} = 400 	imes \left( \frac{27}{8} ight)^{\frac{5}{3}-1} = 400 	imes \left( \frac{27}{8} ight)^{\frac{2}{3}} = 400 	imes \left( \left( \frac{3}{2} ight)^3 ight)^{\frac{2}{3}} = 400 	imes \left( \frac{3}{2} ight)^2 = 400 	imes \frac{9}{4} = 900 \ K$.Work done on the system is $W = -\frac{\mu R (T_2 - T_1)}{\gamma - 1}$.Using $R = 2 	ext{ cal/mol K}$,$W = -\frac{1 	imes 2 	imes (900 - 400)}{\frac{5}{3} - 1} = -\frac{2 	imes 500}{\frac{2}{3}} = -\frac{1000 	imes 3}{2} = -1500 	ext{ cal}$.The magnitude of work done on the system is $1500 	ext{ cal}$.

One mole of an ideal gas $\left(\gamma = \frac{5}{3}\right)$ at $127^{\circ} C$ is compressed adiabatically to $\left(\frac{8}{27}\right)$ of its initial volume. Find the magnitude of the work done on the system in $cal$.

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