Answer
(a) $0.315$
(b) $0.630$
(c) $100KPa;222K$
Work Step by Step
(a) We know that
$P_iV_i^{\gamma}=P_fV_f^{\gamma}$
This can be rearranged as:
$\frac{P_f}{P_i}=(\frac{V_i}{V_f})^{\gamma}$
$\frac{P_f}{P_i}=(\frac{V_i}{2V_i})^{\gamma}$
$\frac{P_f}{P_i}=(\frac{1}{2})^{\frac{5}{3}}$
$\frac{P_f}{P_i}=0.315$
(b) We know that
$\frac{P_iV_i}{T_i}=\frac{P_fV_f}{T_f}$
This can be rearranged as:
$\frac{T_f}{T_i}=(\frac{P_f}{P_i})(\frac{V_f}{V_i})$
We plug in the known values to obtain:
$\frac{T_f}{T_i}=(0.315)(\frac{2V_i}{V_i})$
$\frac{T_f}{T_i}=(0.315)(2)=0.630$
(c) We know that
$P_f=(\frac{V_i}{V_f})^{\gamma}$
We plug in the known values to obtain:
$P_f=(330KPa)(\frac{1.2m^3}{2(1.2m^3)})^{\frac{5}{3}}$
$P_f=100KPa$
Now $\frac{P_f}{P_i}=\frac{100KPa}{330KPa}=0.315$
This is the same answer as found in part(a).
We can find the required temperature as follows:
$T_i=\frac{P_iV_i}{nR}$
$\implies T_i=\frac{(330\times 10^3Pa)(1.2m^3)}{(135mol)(8.31J/mol.K)}$
$T_i=353K$
The final temperature is given as
$T_f=(\frac{P_f}{P_i})(\frac{V_f}{V_i})T_i$
We plug in the known values to obtain:
$T_f=(0.315)(\frac{2V_i}{V_i})T_i$
We plug in the known values to obtain:
$T_f=(0.315)(2)(353K)$
$T_f=222K$
Now $\frac{T_f}{T_i}=\frac{222K}{353K}=0.630$
This is the same answer as found in part(b).
