Two mole of an ideal gas is heated at constant pressure of one atmosphere from `27^(@)C` to `127^(@)C`. If `C_(v,m)=20+10^(-2)"T JK"^(-1).mol^(-1)`, then q and `DeltaU` for the process are respectively:

Options

A. `6362.8J,4700J`
B. `3037.2J,4700J`
C. `7062.8J,5400J`
D. None of these

Explanation

To solve the problem, we need to calculate the heat transfer (q) and the change in internal energy (ΔU) for the given process involving an ideal gas. Let's break down the solution step by step.

### Step 1: Calculate the Work Done (W)
For an ideal gas undergoing a process at constant pressure, the work done (W) can be calculated using the formula:
\[ W = -P \Delta V \]

Using the ideal gas law, we can express the change in volume (ΔV) in terms of temperature:
\[ PV = nRT \]
Thus, the change in volume can be expressed as:
\[ W = -P \left( V_2 - V_1 \right) = -P \left( \frac{nRT_2}{P} - \frac{nRT_1}{P} \right) \]
This simplifies to:
\[ W = -nR(T_2 - T_1) \]

### Step 2: Convert Temperatures to Kelvin
Convert the given temperatures from Celsius to Kelvin:
- \( T_1 = 27^\circ C = 300 \, K \)
- \( T_2 = 127^\circ C = 400 \, K \)

### Step 3: Substitute Values into the Work Formula
Now we can substitute the values into the work formula:
- Number of moles (n) = 2
- R (ideal gas constant) = 8.314 J/(mol·K)
- \( T_2 - T_1 = 400 - 300 = 100 \, K \)

So,
\[ W = -2 \times 8.314 \times 100 \]
\[ W = -1662.8 \, J \]

### Step 4: Calculate the Change in Internal Energy (ΔU)
The change in internal energy (ΔU) can be calculated using the formula:
\[ \Delta U = nC_{V,m} \Delta T \]
Where \( C_{V,m} \) is given as:
\[ C_{V,m} = 20 + 10^{-2}T \]

### Step 5: Integrate to Find ΔU
To find ΔU, we need to integrate \( C_{V,m} \) over the temperature range from \( T_1 \) to \( T_2 \):
\[ \Delta U = n \int_{T_1}^{T_2} C_{V,m} \, dT \]
\[ \Delta U = 2 \int_{300}^{400} \left( 20 + 10^{-2}T \right) dT \]

### Step 6: Perform the Integration
Integrating:
\[ \int (20 + 10^{-2}T) \, dT = 20T + \frac{10^{-2}}{2}T^2 \]
Evaluating from 300 to 400:
\[ \Delta U = 2 \left[ 20(400) + \frac{10^{-2}}{2}(400^2) - \left( 20(300) + \frac{10^{-2}}{2}(300^2) \right) \right] \]

Calculating each term:
1. For \( T = 400 \):
\[ 20(400) = 8000 \]
\[ \frac{10^{-2}}{2}(400^2) = \frac{10^{-2}}{2} \times 160000 = 800 \]

2. For \( T = 300 \):
\[ 20(300) = 6000 \]
\[ \frac{10^{-2}}{2}(300^2) = \frac{10^{-2}}{2} \times 90000 = 450 \]

Now substituting back:
\[ \Delta U = 2 \left[ (8000 + 800) - (6000 + 450) \right] \]
\[ = 2 \left[ 8800 - 6450 \right] \]
\[ = 2 \times 2350 = 4700 \, J \]

### Step 7: Calculate Heat Transfer (q)
Using the first law of thermodynamics:
\[ \Delta U = q + W \]
Rearranging gives:
\[ q = \Delta U - W \]
Substituting the values:
\[ q = 4700 - (-1662.8) \]
\[ q = 4700 + 1662.8 = 6362.8 \, J \]

### Final Results
Thus, the values for q and ΔU are:
- \( q = 6362.8 \, J \)
- \( \Delta U = 4700 \, J \)