Suppose that 5 g of helium gas is heated from `-30^@C` to `120^@C`. Find its change in internal energy and the work it does if the heating occurs (a) at constant volume and (b) at constant pressure. For helium, `C_v`=0.75 cal/g. `C^@` and `C_p` =1.25 cal/g . `C^@`

Answer

To solve the problem of finding the change in internal energy and work done by 5 g of helium gas heated from -30°C to 120°C at both constant volume and constant pressure, we will follow these steps:

### Given Data:
- Mass of helium gas, \( m = 5 \, \text{g} \)
- Initial temperature, \( T_i = -30 \, \text{°C} = -30 + 273.15 = 243.15 \, \text{K} \)
- Final temperature, \( T_f = 120 \, \text{°C} = 120 + 273.15 = 393.15 \, \text{K} \)
- Specific heat at constant volume, \( C_v = 0.75 \, \text{cal/g°C} \)
- Specific heat at constant pressure, \( C_p = 1.25 \, \text{cal/g°C} \)

### Step 1: Calculate the change in temperature (\( \Delta T \))
\[ \Delta T = T_f - T_i = 120 - (-30) = 120 + 30 = 150 \, \text{°C} \]

### Step 2: Case (a) - Heating at Constant Volume
1. **Work Done (\( W \))**:
- At constant volume, the change in volume (\( dV \)) is zero.
\[ W = P \Delta V = 0 \quad \text{(since } \Delta V = 0\text{)} \]

2. **Change in Internal Energy (\( \Delta U \))**:
- The change in internal energy is given by:
\[ \Delta U = m C_v \Delta T \]
- Substituting the values:
\[ \Delta U = 5 \, \text{g} \times 0.75 \, \text{cal/g°C} \times 150 \, \text{°C} = 562.5 \, \text{cal} \]
- Converting calories to joules (1 cal = 4.182 J):
\[ \Delta U = 562.5 \, \text{cal} \times 4.182 \, \text{J/cal} = 2352.375 \, \text{J} \]

### Step 3: Case (b) - Heating at Constant Pressure
1. **Change in Internal Energy (\( \Delta U \))**:
- The change in internal energy remains the same as in the constant volume case:
\[ \Delta U = 562.5 \, \text{cal} \times 4.182 \, \text{J/cal} = 2352.375 \, \text{J} \]

2. **Work Done (\( W \))**:
- The heat added at constant pressure is given by:
\[ Q = m C_p \Delta T \]
- Substituting the values:
\[ Q = 5 \, \text{g} \times 1.25 \, \text{cal/g°C} \times 150 \, \text{°C} = 937.5 \, \text{cal} \]
- Converting to joules:
\[ Q = 937.5 \, \text{cal} \times 4.182 \, \text{J/cal} = 3925.125 \, \text{J} \]

- Using the first law of thermodynamics:
\[ Q = \Delta U + W \implies W = Q - \Delta U \]
- Substituting the values:
\[ W = 3925.125 \, \text{J} - 2352.375 \, \text{J} = 1572.75 \, \text{J} \]

### Final Answers:
- **Case (a)**:
- Change in internal energy, \( \Delta U = 2352.375 \, \text{J} \)
- Work done, \( W = 0 \, \text{J} \)

- **Case (b)**:
- Change in internal energy, \( \Delta U = 2352.375 \, \text{J} \)
- Work done, \( W = 1572.75 \, \text{J} \)