If 8.00 moles of a monatomic ideal gas at a temperature of 260 are expanded isothermally from a volume of 1.08 to a volume of 4.14 .Calculate the work done by the gas.Calculate the heat flow into or out of the gas.If the number of moles is doubled, by what factors do your answers to parts A and B change?
To calculate the work done by the gas, we can use the formula for the work done in an isothermal expansion:
Work = nRT * ln(V2/V1)
Where:
- n is the number of moles of the gas
- R is the ideal gas constant (8.314 J/mol·K)
- T is the temperature of the gas in Kelvin
- V2 is the final volume
- V1 is the initial volume
Let's calculate the work done:
Work = (8.00 mol) * (8.314 J/mol·K) * (260 K) * ln(4.14 L / 1.08 L)
Note that we convert the volumes to liters as the ideal gas constant has SI units.
Calculating this expression gives us:
Work = 8.00 * 8.314 * 260 * ln(4.14 / 1.08)
Work ≈ 8508.566 J (rounded to four decimal places)
Next, let's calculate the heat flow into or out of the gas during the isothermal process. Since the process is isothermal, the temperature remains constant. Therefore, the heat flow (Q) can be calculated using the formula:
Q = nRT * ln(V2/V1)
Using the same values as before, we substitute them into the formula:
Q = (8.00 mol) * (8.314 J/mol·K) * (260 K) * ln(4.14 L / 1.08 L)
Calculating this expression gives us:
Q = 8.00 * 8.314 * 260 * ln(4.14 / 1.08)
Q ≈ 8508.566 J (rounded to four decimal places)
Now, if we double the number of moles of the gas, the values for work (W) and heat flow (Q) will change as follows:
W' = 2 * W (Work is directly proportional to the number of moles)
Q' = 2 * Q (Heat flow is directly proportional to the number of moles)
To calculate the work done by the gas during an isothermal expansion, we need to use the equation:
\[ \text{Work} = -nRT \ln \left( \frac{V_f}{V_i} \right) \]
Where:
- \( n \) is the number of moles of gas
- \( R \) is the ideal gas constant (\( 8.314 \, \text{J/(mol K)} \))
- \( T \) is the temperature in Kelvin
- \( V_f \) is the final volume of the gas
- \( V_i \) is the initial volume of the gas
Given:
\( n = 8.00 \, \text{moles} \)
\( T = 260 \, \text{K} \)
\( V_i = 1.08 \)
\( V_f = 4.14 \)
1. Calculating the work done:
Substituting the given values into the equation, we have:
\[ \text{Work} = - (8.00)(8.314)(260) \ln \left( \frac{4.14}{1.08} \right) \]
Calculating this expression will give the value of the work done by the gas.
2. Calculating the heat flow:
For an isothermal process, the heat flow into or out of the gas (\( Q \)) is equal to the work done (\( W \)) since the internal energy (\( U \)) of the gas remains constant. Therefore, \( Q = W \).
3. Doubling the number of moles:
If the number of moles is doubled from 8.00 to 16.00, the work done would change according to the equation:
\[ \text{New Work} = - (16.00)(8.314)(260) \ln \left( \frac{4.14}{1.08} \right) \]
And the heat flow would also change accordingly:
\[ \text{New Heat Flow} = \text{New Work} \]
By substituting the new number of moles into the equations, you can calculate the new values.