Answer
$124.7 ^{\circ}C$
Work Step by Step
$$PV=nRT$$
$P = 8.27 kPa = 8.27 \times 10^{3}Pa$
$V = 250 cm^{3} = 250 \times 10^{-6} m^{3}$
Find molar mass of $CH_{4} = (12 + 1\times 4)gmol^{-1} = 16gmol^{-1}$
$n = 1.0g/16gmol^{-1}$
$R$ (Universal gas constant) $= 8.3145 JK^{-1}mol^{-1}$
Substitute values in equation to find $T$ in Kelvin,
$T=\frac{PV}{nR}$ = $\frac{ 8.27 \times 10^{3}Pa \times 250 \times 10^{-6} m^{3}}{(1.0g/16gmol^{-1})\times 8.3145 JK^{-1}mol^{-1}} = 397.85 K$
To convert Kelvin to Celsius subtract 273.15,
Temperature in Celsius = $397.85 - 273.15 = 124.7^{\circ}C$
