Final answer:
The work done by 5 moles of an ideal gas expanding from 1 liter to 5 liters at 100°C is calculated using the ideal gas law and the integral of PdV for isothermal expansion. The work done is positive and is found to be approximately 7743.84 J, which is option c. 6.7 × 10³ J.
Step-by-step explanation:
To calculate the work done by 5 moles of an ideal gas undergoing expansion from 1 liter to 5 liters at 100°C, first, we need to apply the ideal gas law to find the pressure of the gas (since it is not provided). The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Since 0.08206 L'atm = 8.314 J and 1 L'atm = 101.3 J, the ideal gas constant R can be used in the form of 8.314 J/mol·K.
To transform the temperature to Kelvin, we use the formula T(K) = T(°C) + 273.15, which gives us T(K) = 100 + 273.15 = 373.15 K. Now we can find the initial pressure P using the initial volume (1 L) with the ideal gas law and n = 5 moles of the gas.
PV = nRT → P = nRT / V
P = (5 moles)(8.314 J/mol·K)(373.15 K) / (1 L) = 15529.6 J/L
Now, we can calculate the work done during the isothermal expansion using the formula W = -P∆V, where ∆V is the change in volume (5 L - 1 L = 4 L), and P is the pressure that we have calculated. The work W is negative because the system is doing work on the surroundings.
W = -P∆V
W = -(15529.6 J/L)(4 L) = -62118.4 J
However, during an isothermal expansion of an ideal gas, the internal pressure changes as the volume changes. Therefore, the correct calculation involves integrating the expression PdV, which is equivalent to nRTln(V_final/V_initial) for an ideal gas undergoing a reversible, isothermal expansion. Note that the work done by the gas will be positive according to this convention.
W = nRTln(V_final/V_initial)
W = (5 moles)(8.314 J/mol·K)(373.15 K)ln(5/1) = 7743.84 J
Thus, the work done by the gas is approximately 7743.84 J, which is closest to the option 6.7 × 103 J (c).
Therefore, the correct answer is c. 6.7 × 103 J.