A diatomic gas (gamma = 1.4$\gamma = 1.4$) does 100 text J$100 \text{ J}$ of work when it is expanded isobarically. Then the heat given to the gas ________ J.

### Related Formula W = PDelta V = nRDelta T$$W = P\Delta V = nR\Delta T$$ Q = nC_pDelta T$$Q = nC_p\Delta T$$ C_p = left(fracf2 + 1right)R$$C_p = \left(\frac{f}{2} + 1\right)R$$ ### Core Logic For an isobaric (constant pressure) process, work done is given by: W = nRDelta T = 100 text J$$W = nR\Delta T = 100 \text{ J}$$ For a diatomic gas, the degrees of freedom f = 5$f = 5$. Thus, the molar heat capacity at constant pressure is: C_p = left(frac52 + 1right)R = frac72R$$C_p = \left(\frac{5}{2} + 1\right)R = \frac{7}{2}R$$ ### Step 1: Calculating Heat Transfer The heat supplied to the gas is: Q = nC_pDelta T = n left(frac72Rright)Delta T = frac72 (nRDelta T)$$Q = nC_p\Delta T = n \left(\frac{7}{2}R\right)\Delta T = \frac{7}{2} (nR\Delta T)$$ ### Step 2: Final Conclusion Substitute the value of work done: Q = frac72 times (100) = 350 text J$$Q = \frac{7}{2} \times (100) = 350 \text{ J}$$ ### Pattern Recognition In an isobaric process, the ratio of Work : Internal Energy Change : Heat Added (W : Delta U : Q$W : \Delta U : Q$) is always 2 : f : (f+2)$2 : f : (f+2)$. For diatomic gases, f=5$f=5$, so the ratio is 2:5:7$2:5:7$. Thus Q = frac72W$Q = \frac{7}{2}W$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics Class 11 Physics: Kinetic Theory of Gases