If three moles of an ideal gas at 300 K expand isothermally from 30 dm3 to 45 dm3 against a constant opposing pressure of 80 kPa, then the amount of heat transferred is ______ J.
Correct Answer: 1200
Approach Solution - 1
To find the heat transferred during the isothermal expansion of an ideal gas, we use the formula for work done in isothermal expansion: \( W = -P_{\text{ext}} \times \Delta V \). Here, \( P_{\text{ext}} = 80 \) kPa or \( 80,000 \) Pa, \(\Delta V = V_f - V_i = 45 \) dm3 - 30 dm3 = 15 dm3 or \( 0.015 \) m3.
Plugging in the values:
\( W = -80,000 \times 0.015 = -1,200 \) J.
The first law of thermodynamics states \( \Delta U = Q + W \). For isothermal processes, \( \Delta U = 0 \) because the internal energy change \(\Delta U\) is zero (temperature is constant). Thus, \( 0 = Q + W \) or \( Q = -W \).
Therefore, the heat transferred \( Q = 1,200 \) J.
Approach Solution -2
Using the first law of thermodynamics:
\[ \Delta U = Q + W \]
For an isothermal process, \(\Delta U = 0\), so \(Q = -W\).
\[ W = -P_{\text{ext}} \Delta V = -80 \times 10^3 \times (45 - 30) \times 10^{-3} = -1200 \, \text{J} \]
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