1. Use the relation between degrees of freedom and molar heat capacities

For an ideal gas with degrees of freedom $f$:

$C_V = \frac{f}{2}R$

and

$C_P = C_V + R = \left(\frac{f}{2}+1\right)R = \frac{f+2}{2}R$

Given:

$f = 8$

So,

$C_P = \frac{8+2}{2}R = 5R$

2. Use the work done in constant-pressure expansion

For constant pressure,

$W = P\Delta V = nR\Delta T$

Given:

$W = 150\,\text{J}$

Hence,

$nR\Delta T = 150$

3. Compute heat absorbed at constant pressure

At constant pressure,

$Q = nC_P\Delta T$

Substitute $C_P = 5R$:

$Q = n(5R)\Delta T = 5(nR\Delta T)$

Using $nR\Delta T = 150$:

$Q = 5 \times 150 = 750\,\text{J}$

4. Final Answer

$\boxed{750}$

5. Comparison with stored answer

Stored correct answer = $750$

This matches the derived answer.