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Step-by-Step Solution
Step 1: Identify the Key Physical Quantities
• We have a cylinder of volume 67.2 L containing helium gas.
• The process is carried out at a constant volume (since the cylinder has a fixed capacity).
• The temperature of the gas needs to be raised by 20 °C.
• For a monoatomic ideal gas (like helium), the molar heat capacity at constant volume is
$C_v = \frac{3}{2}R$.
• We are given the gas constant $R = 8.31\text{ J mol}^{-1}\text{K}^{-1}$.
Step 2: Calculate Number of Moles of Helium
At STP (standard temperature and pressure), 1 mole of an ideal gas occupies 22.4 L. The volume given is 67.2 L, so
$$
n = \frac{\text{Volume at STP}}{\text{Molar volume at STP}}
= \frac{67.2}{22.4}
= 3 \text{ moles}.
$$
Step 3: Use the Formula for Heat Required at Constant Volume
The heat required $ \Delta Q $ to raise the temperature of $n$ moles of a gas by $\Delta T$ at constant volume is
$$
\Delta Q = n \, C_{v} \, \Delta T.
$$
For a monoatomic gas, $ C_{v} = \frac{3}{2} R $. Thus,
$$
\Delta Q = n \left(\frac{3}{2} R\right) \Delta T.
$$
Step 4: Substitute the Values
Here, $n = 3$, $R = 8.31\text{ J mol}^{-1}\text{K}^{-1}$, and $\Delta T = 20\,^\circ\mathrm{C} = 20\,\mathrm{K}$ (the increment in °C and K is numerically the same):
$$
\Delta Q
= 3
\times \left(\frac{3}{2} \times 8.31 \right)
\times 20.
$$
Simplifying:
$$
\Delta Q
= 3
\times \frac{3}{2}
\times 8.31
\times 20.
$$
Step 5: Numerical Evaluation
First calculate the factor $ \frac{3}{2} \times 8.31 = 12.465 $. Then,
$$
\Delta Q
= 3 \times 12.465 \times 20
\approx 3 \times 249.3
= 747.9 \approx 748 \,\text{J}.
$$
Result
Therefore, the amount of heat needed to raise the temperature of the helium gas by 20 °C is approximately
748 J.
