Answer
(a) This is an isobaric process.
(b) The final temperature is $118^{\circ}C$
(c) There are $6.23\times 10^{-3}$ moles of gas.
Work Step by Step
(a) The pressure remains constant so this is an isobaric process.
(b) We can convert the initial temperature to Kelvin.
$K = C + 273$
$K = 900 + 273$
$K = 1173$
The initial temperature is 1173 K.
We can find an expression for the initial temperature.
$PV_1 = nRT_1$
$T_1 = \frac{PV_1}{nR}$
We can find the final temperature.
$PV_2 = nRT_2$
$T_2 = \frac{PV_2}{nR}$
$T_2 = \frac{P(V_1/3)}{nR}$
$T_2 = \frac{1}{3}~\frac{PV_1}{nR}$
$T_2 = \frac{T_1}{3}$
$T_2 = \frac{1173~K}{3}$
$T_2 = 391~K$
We can convert the final temperature to Celsius.
$C = K - 273$
$C = 391-273$
$C = 118$
The final temperature is $118^{\circ}C$
(c) We can use the initial conditions to find the number of moles of gas.
$PV = nRT$
$n = \frac{PV}{RT}$
$n = \frac{(2)(1.013\times 10^5~Pa)(300\times 10^{-6}~m^3)}{(8.314~m^3~Pa/mol~K)(1173~K)}$
$n = 6.23\times 10^{-3}~mol$
There are $6.23\times 10^{-3}$ moles of gas.
