**Problem:**A sample of 2.00 mol of a perfect gas is initially at a pressure of 111 kPa and temperature of 277 K. The sample is heated reversibly to 356 K at a constant volume. The constant volume molar heat capacity is 2.5R. Calculate the final pressure, ΔU, q, and w.**Solution:**1. **Given Data:** - Number of moles (n): 2.00 mol - Initial temperature (T₁): 277 K - Final temperature (T₂): 356 K - Pressure (P): 111 kPa = 1.11×10⁵ Pa - Change in temperature (ΔT): \( T₂ - T₁ = 356 \, \text{K} - 277 \, \text{K} = 79 \, \text{K} \) - Constant volume molar heat capacity (\( C_{v,m} \)): 2.5R - Molar gas constant (R): 8.314 J/mol·K2. **Calculations:** - \( C_{v,m} = 2.5 \times 8.314 \, \text{J/mol·K} = 20.79 \, \text{J/mol·K} \)3. **Key Assumptions:** - Work done (w): 0 (constant volume process) - \(\Delta U = q\) (since \( w = 0 \))4. **Notes:** - A reversible process indicates that the pressure inside is the same as the pressure outside at equilibrium. - Energy transfer as heat (q) is calculated using the formula \( q = m \cdot C_s \cdot \Delta T \).This outlines the problem setup, given conditions, and preliminary calculations necessary to solve for the final pressure, change in internal energy (\(\Delta U\)), heat transfer (q), and work done (w).

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5.) A sample of 2.00 mol of a perfect gas is initially at a pressure of 111 kPa and temperature of 277 K. The sample is heated reversibly to 356 K at a constant volume. The constant volume molar heat capacity is 2.5R. Calculate the final pressure, AU, q, and w.

5.) A sample of 2.00 mol of a perfect gas is initially at a pressure of 111 kPa and temperature of 277 K. The sample is heated reversibly to 356 K at a constant volume. The constant volume molar heat capacity is 2.5R. Calculate the final pressure, AU, q, and w.

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Chapter1: Chemical Foundations

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Question

**Problem:**
A sample of 2.00 mol of a perfect gas is initially at a pressure of 111 kPa and temperature of 277 K. The sample is heated reversibly to 356 K at a constant volume. The constant volume molar heat capacity is 2.5R. Calculate the final pressure, ΔU, q, and w.

**Solution:**

1. **Given Data:**
- Number of moles (n): 2.00 mol
- Initial temperature (T₁): 277 K
- Final temperature (T₂): 356 K
- Pressure (P): 111 kPa = 1.11×10⁵ Pa
- Change in temperature (ΔT): \( T₂ - T₁ = 356 \, \text{K} - 277 \, \text{K} = 79 \, \text{K} \)
- Constant volume molar heat capacity (\( C_{v,m} \)): 2.5R
- Molar gas constant (R): 8.314 J/mol·K

2. **Calculations:**
- \( C_{v,m} = 2.5 \times 8.314 \, \text{J/mol·K} = 20.79 \, \text{J/mol·K} \)

3. **Key Assumptions:**
- Work done (w): 0 (constant volume process)
- \(\Delta U = q\) (since \( w = 0 \))

4. **Notes:**
- A reversible process indicates that the pressure inside is the same as the pressure outside at equilibrium.
- Energy transfer as heat (q) is calculated using the formula \( q = m \cdot C_s \cdot \Delta T \).

This outlines the problem setup, given conditions, and preliminary calculations necessary to solve for the final pressure, change in internal energy (\(\Delta U\)), heat transfer (q), and work done (w).

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