$A$ cylinder of fixed capacity $44.8 \, L$ contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by $15.0 \, ^{\circ}C$? $(R = 8.31 \, J \, mol^{-1} K^{-1})$

(C) Using the ideal gas law $PV = \mu RT$,we know that $1 \, mol$ of any ideal gas at standard temperature $(273 \, K)$ and pressure $(1 \, atm = 1.01 \times 10^5 \, Pa)$ occupies a volume of $22.4 \, L$.
Since the cylinder has a volume of $44.8 \, L$,the number of moles of helium is $\mu = \frac{44.8}{22.4} = 2 \, mol$.
Helium is a monatomic gas,so its molar specific heat at constant volume is $C_v = \frac{3}{2}R$.
Since the volume of the cylinder is fixed,the process is isochoric,and the heat required is given by $Q = \mu C_v \Delta T$.
Substituting the values: $Q = 2 \times (\frac{3}{2} \times 8.31) \times 15.0$.
$Q = 3 \times 8.31 \times 15.0 = 45 \times 8.31 = 373.95 \, J \approx 374 \, J$.