In this problem, we are tasked with calculating the total internal energy of an ideal monoatomic gas contained in a tank. The internal energy, denoted as \( E_{\text{internal}} \), can be determined using the equation:
\( E_{\text{internal}} = \frac{3}{2} nRT \)
Here, \( n \) represents the number of moles, \( R \) is the ideal gas constant (approximately \( 8.314 \, \text{J/(mol·K)} \)), and \( T \) is the temperature in Kelvin. To find \( E_{\text{internal}} \), we need to know two of the three variables in the equation. In this case, we are given \( n = 10 \) moles, but we need to determine the temperature \( T \).
Since the temperature is not provided directly, we can use the ideal gas law, expressed as:
\( PV = nRT \)
From this equation, we can solve for temperature \( T \) as follows:
\( T = \frac{PV}{nR} \)
Given the pressure \( P = 0.8 \, \text{atm} \) and volume \( V = 0.3 \, \text{m}^3 \), we first convert the pressure from atmospheres to Pascals using the conversion factor \( 1 \, \text{atm} = 1.01 \times 10^5 \, \text{Pa} \):
\( P = 0.8 \, \text{atm} \times 1.01 \times 10^5 \, \text{Pa/atm} = 8.08 \times 10^4 \, \text{Pa} \)
Now, substituting the values into the equation for temperature:
\( T = \frac{(8.08 \times 10^4 \, \text{Pa})(0.3 \, \text{m}^3)}{(10 \, \text{mol})(8.314 \, \text{J/(mol·K)})} \approx 291.6 \, \text{K} \)
With the temperature calculated, we can now substitute \( n \), \( R \), and \( T \) back into the internal energy equation:
\( E_{\text{internal}} = \frac{3}{2} (10)(8.314)(291.6) \approx 3.64 \times 10^4 \, \text{J} \)
Alternatively, we can derive the internal energy directly using the relationship between the ideal gas law and the internal energy equation. By substituting \( nRT \) with \( PV \) in the internal energy equation, we have:
\( E_{\text{internal}} = \frac{3}{2} PV \)
Substituting the known values:
\( E_{\text{internal}} = \frac{3}{2} (8.08 \times 10^4 \, \text{Pa})(0.3 \, \text{m}^3) \approx 3.64 \times 10^4 \, \text{J} \)
Both methods yield the same result, confirming the internal energy of the gas in the tank is approximately \( 3.64 \times 10^4 \, \text{J} \).