4 moles of a monoatomic gas are filled in a rigid container. Temperature of the gas increases from `27^(@)C` to `127^(@)C` by heating. Let Q, `DeltaU` and W represent heat, change in internal energy and work done, then

Options

A. Q=600R
B. Q=1000R
C. `DeltaU=600R`
D. `DeltaU=800R`

Answer

To solve the problem, we need to find the values of heat (Q), change in internal energy (ΔU), and work done (W) for the given monoatomic gas in a rigid container.

### Step 1: Identify the given values
- Number of moles (n) = 4 moles
- Initial temperature (T1) = 27°C = 300 K (convert to Kelvin by adding 273)
- Final temperature (T2) = 127°C = 400 K (convert to Kelvin by adding 273)

### Step 2: Calculate the change in temperature (ΔT)
\[ \Delta T = T2 - T1 = 400 K - 300 K = 100 K \]

### Step 3: Determine the specific heat at constant volume (Cv) for a monoatomic gas
For a monoatomic ideal gas, the specific heat at constant volume is given by:
\[ C_v = \frac{3}{2} R \]
where R is the universal gas constant.

### Step 4: Calculate the change in internal energy (ΔU)
The change in internal energy for an ideal gas is given by the formula:
\[ \Delta U = n C_v \Delta T \]
Substituting the values:
\[ \Delta U = 4 \cdot \left(\frac{3}{2} R\right) \cdot 100 \]
\[ \Delta U = 4 \cdot \frac{3}{2} \cdot 100 R = 600 R \]

### Step 5: Determine the work done (W)
Since the gas is in a rigid container, the volume does not change, which means:
\[ W = 0 \]

### Step 6: Apply the first law of thermodynamics
The first law of thermodynamics states:
\[ Q = \Delta U + W \]
Substituting the values we have:
\[ Q = 600 R + 0 = 600 R \]

### Conclusion
Thus, we find that:
- Heat added (Q) = 600 R
- Change in internal energy (ΔU) = 600 R
- Work done (W) = 0

### Final Answer
Both Q and ΔU are equal to 600 R, and W is 0.

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