To calculate the work done on compressing an ideal gas at constant temperature, we can use the formula for work done during isothermal compression:
\[ W = -P_{V} \ln\left(\frac{P_2}{P_1}\right) \]
Where:
- \(W\) is the work done on the gas,
- \(P_1\) is the initial pressure,
- \(P_2\) is the final pressure,
- \(P_V\) is the specific volume.
Because we need the specific volume, we can calculate it using the ideal gas equation:
\[ PV = nRT \implies V = \frac{nRT}{P} \]
First, we need to determine the number of moles (n) of gas using the given mass and the specific gas constant. Since R is given in kJ/kg.K, we can find the specific volume for both pressures using the mass of the gas.
- Calculate the number of moles (n):
Given:
- Mass (m) = 4.54 kg
- Gas constant (R) = 0.3 kJ/kg.K
- Temperature (T) = 432 K
The number of moles (n) can be calculated as follows:
From the ideal gas constant, \(R\):
\[ R = \frac{R_u}{M} \] Where,
- \(R_u\) = universal gas constant (approximately \(8.314 , \text{J/mol.K}\))
- M = molar mass in kg/mol
Since we can directly use the specific gas constant for calculations, we will continue with the calculations using the initial and final pressure values.
- Calculate Work Done (W): We can rewrite the work done as:
\[ W = -mRT \ln\left(\frac{P_2}{P_1}\right) \]
Where:
- \(m\) is the mass,
- \(R\) is the specific gas constant,
- \(T\) is the absolute temperature,
- \(P_1 = 103.4 , \text{kPa} = 103400 , \text{Pa}\)
- \(P_2 = 3 \times 102 , \text{kPa} = 310200 , \text{Pa}\)
Now we can substitute the values into the equation:
\[ W = -4.54 \times 0.3 \times 432 \ln\left(\frac{310200}{103400}\right) \]
- Calculate \(\ln\left(\frac{310200}{103400}\right)\):
\[ \frac{310200}{103400} \approx 3.00 \] \[ \ln(3.00) \approx 1.0986 \]
- Substitute back into the work equation:
\[ W = -4.54 \times 0.3 \times 432 \times 1.0986 \] \[ W = -4.54 \times 0.3 \times 432 \times 1.0986 \approx -54.80 , \text{kJ} \]
So the work done on compressing the gas is approximately:
\[ \boxed{-54.80 , \text{kJ}} \]
(Note: The negative sign indicates work is done on the gas, which is conventionally represented as a negative value in thermodynamic calculations.)