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Step-by-Step Solution
Step 1: Identify the relevant principle
We use the First Law of Thermodynamics, which states:
$$\Delta Q = \Delta U + \Delta W,$$
where
- $\Delta Q$ is the heat added to the system,
- $\Delta U$ is the change in internal energy,
- $\Delta W$ is the work done by the system.
Step 2: Note the condition "no work is done"
Given that no work is done on or by the gas,
$$\Delta W = 0.$$
Hence, from the relation
$$\Delta Q = \Delta U + \Delta W,$$
we get
$$\Delta Q = \Delta U.$$
Step 3: Express the change in internal energy for a diatomic gas
The change in internal energy for $n$ moles of a diatomic gas at constant volume is given by:
$$\Delta U = n C_V \Delta T,$$
where $C_V$ is the molar heat capacity at constant volume for a diatomic gas.
Step 4: Substitute the value of $C_V$ for a diatomic gas
For a rigid (non-vibrating) diatomic gas, the molar heat capacity at constant volume ($C_V$) is:
$$C_V = \frac{5R}{2}.$$
Step 5: Insert all given data
- Number of moles, $n = 4.$
- Temperature change, $\Delta T = 50^\circ\text{C} - 0^\circ\text{C} = 50.$
- $C_V = \frac{5R}{2}.$
Thus,
$$\Delta U = n C_V \Delta T = 4 \times \frac{5R}{2} \times 50.$$
Step 6: Calculate the final value of heat ($\Delta Q$)
Since $\Delta Q = \Delta U$,
\[
\Delta Q = 4 \times \frac{5R}{2} \times 50 = 4 \times \frac{5}{2} \times 50 \; R = 4 \times \frac{5}{2} \times 50 \; R.
\]
First simplify the constants:
\[
4 \times \frac{5}{2} = 10, \quad 10 \times 50 = 500.
\]
So,
\[
\Delta Q = 500 R.
\]
Step 7: State the final answer
The amount of heat needed to raise the temperature of the gas under the given conditions is
$$\boxed{500 R}.$$
