5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K
If `C_(V) = 28 J K^(-1)mol^(-1)`, calculate `DeltaU` and `DeltapV` for this process. (R = 8.0 `JK^(-1)mol^(-1)`]

Options

A. `DeltaU = 14 kJ, Delta(pV) = 4 kJ`
B. `DeltaU = 14 kJ, Delta(pV) = 18 kJ`
C. `DeltaU = 2.8 kJ, Delta(pV) = 0.8 kJ`
D. `DeltaU = 14 kJ, Delta(pV) = 0.4 kJ`

Explanation

To solve the problem, we will calculate the change in internal energy (ΔU) and the change in pressure-volume work (ΔPV) for the given process of an ideal gas.

### Step 1: Calculate ΔU (Change in Internal Energy)

The formula for the change in internal energy for an ideal gas is given by:

\[ \Delta U = N C_V \Delta T \]

Where:
- \(N\) = number of moles = 5 moles
- \(C_V\) = heat capacity at constant volume = 28 J K\(^{-1}\) mol\(^{-1}\)
- \(\Delta T\) = change in temperature = \(T_2 - T_1 = 200 K - 100 K = 100 K\)

Now substituting the values into the formula:

\[ \Delta U = 5 \, \text{moles} \times 28 \, \text{J K}^{-1} \text{mol}^{-1} \times 100 \, \text{K} \]

\[ \Delta U = 5 \times 28 \times 100 = 14000 \, \text{J} \]

Converting to kilojoules:

\[ \Delta U = 14 \, \text{kJ} \]

### Step 2: Calculate ΔH (Change in Enthalpy)

The change in enthalpy (ΔH) can be calculated using the formula:

\[ \Delta H = N C_P \Delta T \]

Where \(C_P\) can be calculated using the relation:

\[ C_P = C_V + R \]

Given \(R = 8.0 \, \text{J K}^{-1} \text{mol}^{-1}\):

\[ C_P = 28 \, \text{J K}^{-1} \text{mol}^{-1} + 8.0 \, \text{J K}^{-1} \text{mol}^{-1} = 36 \, \text{J K}^{-1} \text{mol}^{-1} \]

Now substituting into the enthalpy formula:

\[ \Delta H = 5 \, \text{moles} \times 36 \, \text{J K}^{-1} \text{mol}^{-1} \times 100 \, \text{K} \]

\[ \Delta H = 5 \times 36 \times 100 = 18000 \, \text{J} \]

Converting to kilojoules:

\[ \Delta H = 18 \, \text{kJ} \]

### Step 3: Calculate ΔPV

Using the relationship between ΔH, ΔU, and ΔPV:

\[ \Delta H = \Delta U + \Delta PV \]

Rearranging gives:

\[ \Delta PV = \Delta H - \Delta U \]

Substituting the values we found:

\[ \Delta PV = 18 \, \text{kJ} - 14 \, \text{kJ} = 4 \, \text{kJ} \]

### Final Results

- \(\Delta U = 14 \, \text{kJ}\)
- \(\Delta PV = 4 \, \text{kJ}\)

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