For an isobaric process, the work done is:
\[w = P \Delta V = nR \Delta T.\]
Given:
\[w = 100 \, \text{J}.\]
The heat (\(Q\)) supplied is:
\[Q = \Delta U + w,\]
where the internal energy change (\(\Delta U\)) for a diatomic gas is:
\[\Delta U = \frac{f}{2} nR \Delta T,\]
with \(f = 5\) (degrees of freedom for a diatomic gas).
Substitute:
\[Q = \frac{f}{2} nR \Delta T + nR \Delta T.\]
Simplify:
\[Q = \left(\frac{f}{2} + 1\right) nR \Delta T.\]
Substitute \(\Delta T\) from \(w = nR \Delta T\):
\[Q = \left(\frac{f}{2} + 1\right) \times w.\]
Substitute \(f = 5\) and \(w = 100 \, \text{J}\):
\[Q = \left(\frac{5}{2} + 1\right) \times 100 = \left(\frac{7}{2}\right) \times 100 = 350 \, \text{J}.\]
Thus, the heat given to the gas is:
\[Q = 350 \, \text{J}.\]
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to: