Answer
(a) This is an isochoric process.
(b) The initial temperature is $1555^{\circ}C$
The final temperature is $336^{\circ}C$
Work Step by Step
(a) Since the volume does not change, this is an isochoric process.
(b) We can find the initial temperature as:
$P_1V = nRT_1$
$T_1 = \frac{P_1V}{nR}$
$T_1 = \frac{(3)(1.013\times 10^5~Pa)(200\times 10^{-6}~m^3)}{(0.0040~mol)(8.314~m^3~Pa/mol~K)}$
$T_1 = 1828~K$
We then convert this temperature to Celsius:
$C = K - 273$
$C = 1828-273$
$C = 1555$
The initial temperature is $1555^{\circ}C$.
Next, we find the final temperature:
$P_2V = nRT_2$
$T_2 = \frac{P_2V}{nR}$
$T_2 = \frac{(1)(1.013\times 10^5~Pa)(200\times 10^{-6}~m^3)}{(0.0040~mol)(8.314~m^3~Pa/mol~K)}$
$T_2 = 609~K$
We can convert this temperature to Celsius.
$C = K - 273$
$C = 609-273$
$C = 336$
The final temperature is $336^{\circ}C$.
