If three moles of an ideal gas at 300text K$300\text{ K}$ expand isothermally from 30text dm^3$30\text{ dm}^3$ to 45text dm^3$45\text{ dm}^3$ against a constant opposing pressure of 80text kPa$80\text{ kPa}$, then the amount of heat transferred is textquadquad$\text{\quad\quad}$ J.

### Related Formula First law of thermodynamics framework: Delta U = Q + W$$\Delta U = Q + W$$ For an isothermal processes involving ideal gases, internal energy change is zero: Delta U = 0 implies Q = -W$$\Delta U = 0 \implies Q = -W$$ Irreversible work formula expanding against constant external pressure: W = -P_textext Delta V = -P_textext(V_2 - V_1)$$W = -P_{\text{ext}} \Delta V = -P_{\text{ext}}(V_2 - V_1)$$ ### Step 1: Calculate structural work values Given values: P_textext = 80text kPa = 80 times 10^3text Pa$$P_{\text{ext}} = 80\text{ kPa} = 80 \times 10^3\text{ Pa}$$ V_1 = 30text dm^3 = 30 times 10^-3text m^3$$V_1 = 30\text{ dm}^3 = 30 \times 10^{-3}\text{ m}^3$$ V_2 = 45text dm^3 = 45 times 10^-3text m^3$$V_2 = 45\text{ dm}^3 = 45 \times 10^{-3}\text{ m}^3$$ Delta V = (45 - 30) times 10^-3 = 15 times 10^-3text m^3$$\Delta V = (45 - 30) \times 10^{-3} = 15 \times 10^{-3}\text{ m}^3$$ W = -80 times 10^3 times (15 times 10^-3) = -1200text J$$W = -80 \times 10^3 \times (15 \times 10^{-3}) = -1200\text{ J}$$ ### Step 2: Solve for heat magnitude $Q = -W = -(-1200text J) = 1200text J$$$Q = -W = -(-1200\text{ J}) = 1200\text{ J}$$ ### Pattern Recognition Constant opposing pressure indicates an irreversible process path. Use $ ### Pattern Recognition Constant opposing pressure indicates an irreversible process path. Use $W = -P\Delta V$ instead of logarithmic integrals. ### Chapter Mix Class 11 Chemistry: Thermodynamics