A diatomic gas ( y= 1.4) does 100 J of work when it is expanded isobarically. Then the heat given to the gas ____ J.

A diatomic gas with $$\gamma = 1.4$$ does 100 J of work during isobaric (constant pressure) expansion. We need to find the heat given to the gas.

We first recall the key thermodynamic relations for an isobaric process. For an ideal gas at constant pressure:

- Work done: $$W = P\Delta V = nR\Delta T$$ (using the ideal gas law $$PV = nRT$$)

- Heat absorbed: $$Q = nC_p\Delta T$$ (at constant pressure, heat = $$nC_p\Delta T$$)

- The molar heat capacity at constant pressure: $$C_p = \frac{\gamma R}{\gamma - 1}$$ (derived from $$\gamma = \frac{C_p}{C_v}$$ and $$C_p - C_v = R$$)

Next, we find the ratio $$\frac{Q}{W}$$:

$$ \frac{Q}{W} = \frac{nC_p\Delta T}{nR\Delta T} = \frac{C_p}{R} $$

Substituting $$C_p = \frac{\gamma R}{\gamma - 1}$$ gives

$$ \frac{Q}{W} = \frac{\gamma R}{(\gamma - 1) \times R} = \frac{\gamma}{\gamma - 1} $$

Substituting $$\gamma = 1.4$$, we get

$$ \frac{Q}{W} = \frac{1.4}{1.4 - 1} = \frac{1.4}{0.4} = \frac{14}{4} = 3.5 $$

Therefore, the heat absorbed is

$$ Q = 3.5 \times W = 3.5 \times 100 = 350 \text{ J} $$

We can verify this result using the First Law of Thermodynamics. According to the First Law, $$Q = \Delta U + W$$, where $$\Delta U = nC_v\Delta T$$.

We find

$$\frac{\Delta U}{W} = \frac{nC_v\Delta T}{nR\Delta T} = \frac{C_v}{R} = \frac{1}{\gamma - 1} = \frac{1}{0.4} = 2.5$$

Hence, $$\Delta U = 2.5 \times 100 = 250$$ J, and so $$Q = 250 + 100 = 350$$ J, which confirms the previous calculation.

The heat given to the gas is $$\boxed{350}$$ J.