An ideal gas having initial pressure P,volume | vedclass

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(B) For an adiabatic process,the relation between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Substituting the given

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(B) For an adiabatic process,the relation between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Substituting the given values: $T V^{\gamma-1} = (T/2) (4V)^{\gamma-1}$.Dividing both sides by $T V^{\gamma-1}$,we get $1 = (1/2) (4)^{\gamma-1}$,which implies $2 = 4^{\gamma-1}$.Since $4 = 2^2$,we have $2^1 = (2^2)^{\gamma-1} = 2^{2\gamma-2}$.Equating the exponents: $1 = 2\gamma - 2$,so $2\gamma = 3$,which gives $\gamma = 3/2$.The work done in an adiabatic process is $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.Substituting $T_1 = T$,$T_2 = T/2$,and $\gamma = 3/2$: $W = \frac{nR(T - T/2)}{3/2 - 1} = \frac{nR(T/2)}{1/2} = nRT$.Since $PV = nRT$ for an ideal gas,$W = PV$.Comparing $W = \alpha PV$ with $W = PV$,we get $\alpha = 1$.

An ideal gas having initial pressure $P$,volume $V$ and temperature $T$ is allowed to expand adiabatically until its volume becomes $4V$,while its temperature falls to $T/2$. If the work done by the gas during the expansion is $\alpha PV$,the value of $\alpha$ is

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