Physics Thermodynamics

$\mathrm{\Delta }\mathrm{Q}={\mathrm{n}\mathrm{C}}_{\mathrm{p}}\mathrm{\Delta }\mathrm{T}\Rightarrow 160={\mathrm{n}\mathrm{C}}_{\mathrm{p}}50$

$\mathrm{\Delta }Q={\mathrm{n}\mathrm{C}}_{\mathrm{v}}\mathrm{\Delta }\mathrm{T}\Rightarrow 240={\mathrm{n}\mathrm{C}}_{\mathrm{v}}100$

$\begin{array}{rr}& \Rightarrow \mathrm{}\frac{C_p}{C_v}=\frac{16}{5}\times \frac{10}{24}=\frac{4}{3}\\ & \Rightarrow \mathrm{}1+\frac{2}{f}=\frac{4}{3}\mathrm{}\Rightarrow f=6\end{array}$

$m\left(L\right)=m_1{S}_1\left(\mathrm{\Delta }T\right)$

$\Rightarrow \mathrm{m}\left(3.4\times {10}^5\right)=\left(200\right)\left(4200\right)\left(25\right)$

$\Rightarrow \mathrm{m}=61.7$

$\frac{50-40}{300}=-k\left. [\frac{50+40}{2}-20\right]$

$\begin{array}{rr}& \frac{40-\mathrm{T}}{300}-\mathrm{k}\left. [\frac{\mathrm{T}+40}{2}-20\right]\\ & \Rightarrow \mathrm{}\frac{10}{40-\mathrm{T}}=\frac{25}{\mathrm{T}}\times 2\\ & \Rightarrow \mathrm{}\mathrm{T}=200-5\mathrm{T}\\ & \Rightarrow \mathrm{}\mathrm{T}=\frac{100}{3}\approx {33}^{?}\mathrm{C}\end{array}$

Heat lost by steam $=$  Heat gained by water and calorimeter.

$\begin{array}{rr}& \mathrm{m}\times 540+\mathrm{m}\times 1\times (100-31)=200\\ & 540\mathrm{m}+69\mathrm{m}=1200\\ & \mathrm{m}=\frac{1200}{609}\approx 2\end{array}$

After burning, heat exchange occurs between helium and atmospheric. Hence, irreversible, isothermal process.

The internal energy of one mole of an ideal gas is given by U = (f/2)RT, where f is the number of degrees of freedom.
For a non-linear triatomic molecule (assumed to be a rigid rotator), there are:

3 translational degrees of freedom.

3 rotational degrees of freedom.
Vibrational modes are generally not c