Answer:
a) [tex]U = 652.545\,kJ[/tex], b) [tex]v \approx 659.568\,\frac{m}{s}[/tex]
Explanation:
a) According to the First Law of Thermodinamics, the system is not reporting any work, mass or heat interactions. Besides, let consider that such box is rigid and, therefore, heat contained inside is the consequence of internal energy.
[tex]Q = U[/tex]
The internal energy for a monoatomic ideal gas is:
[tex]U = \frac{3}{2} \cdot n \cdot R_{u} \cdot T[/tex]
Let assume that cubical box contains just one kilomole of monoatomic gas. Then, the temperature is determined from the Equation of State for Ideal Gases:
[tex]T = \frac{P\cdot V}{n\cdot R_{u}}[/tex]
[tex]T = \frac{(202.65\,kPa)\cdot(1.29\,m)^{3}}{(1\,kmole)\cdot(8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K} )}[/tex]
[tex]T = 52.325\,K[/tex]
The thermal energy contained by the gas is:
[tex]U = \frac{3}{2}\cdot (1\,kmole)\cdot (8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K})\cdot (52.325\,K)[/tex]
[tex]U = 652.545\,kJ[/tex]
b) The physical model for the cat is constructed from Work-Energy Theorem:
[tex]U = \frac{1}{2}\cdot m_{cat} \cdot v^{2}[/tex]
The speed of the cat is obtained by isolating the respective variable and the replacement of every known variable by numerical values:
[tex]v = \sqrt{\frac{2 \cdot U}{m_{cat}}}[/tex]
[tex]v = \sqrt{\frac{2\cdot (652.545 \times 10^{3}\,J)}{3\,kg} }[/tex]
[tex]v \approx 659.568\,\frac{m}{s}[/tex]
