5 mole of ideal gas at `100K(C_(v.m)=28J//mol//K)`. It is heated upto 200 K. Calculate `triangleU` and `triangle(PV)` for the process `(R=8J//mol-K)`

Options

A. `triangleU=28KJ,triangle(PV)=8KJ`
B. `triangleU=14KJ,triangle(PV)=4KJ`
C. `triangleU=14KJ:triangle(PV)=8KJ`
D. `triangleU=28KJ,triangle(PV)=4KJ`

Answer

To solve the problem, we need to calculate the change in internal energy (ΔU) and the change in (PV) (Δ(PV)) for an ideal gas. Let's break this down step by step.

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

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

\[ \Delta U = n C_{v,m} \Delta T \]

Where:
- \( n \) = number of moles of the gas = 5 moles
- \( C_{v,m} \) = molar heat capacity at constant volume = 28 J/(mol·K)
- \( \Delta T \) = change in temperature = \( T_{final} - T_{initial} = 200 K - 100 K = 100 K \)

Now substituting the values into the formula:

\[ \Delta U = 5 \, \text{moles} \times 28 \, \text{J/(mol·K)} \times 100 \, \text{K} \]

Calculating this gives:

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

Converting joules to kilojoules:

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

### Step 2: Calculate Δ(PV)

The formula for the change in (PV) for an ideal gas is given by:

\[ \Delta(PV) = nR\Delta T \]

Where:
- \( R \) = ideal gas constant = 8 J/(mol·K)
- \( n \) = number of moles of the gas = 5 moles
- \( \Delta T \) = change in temperature = 100 K (as calculated before)

Now substituting the values into the formula:

\[ \Delta(PV) = 5 \, \text{moles} \times 8 \, \text{J/(mol·K)} \times 100 \, \text{K} \]

Calculating this gives:

\[ \Delta(PV) = 5 \times 8 \times 100 = 4000 \, \text{J} \]

Converting joules to kilojoules:

\[ \Delta(PV) = 4 \, \text{kJ} \]

### Final Results

- ΔU = 14 kJ
- Δ(PV) = 4 kJ