Question

A cubical box measuring 1.29 m on each side contains a monatomic ideal gas at a pressure of 2.0 atm How much thermal energy do the particles of this gas contain? Express your answer with the appropriate units. If half the thermal energy of the particles were converted to the kinetic energy of a 3.0-kg cat. how fast would the cat be moving? Express your answer with the appropriate units.

Answer 1

Answer:

a) $U = 652.545\,kJ$, b) $v \approx 659.568\,\frac{m}{s}$

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.

$Q = U$

The internal energy for a monoatomic ideal gas is:

$U = \frac{3}{2} \cdot n \cdot R_{u} \cdot T$

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:

$T = \frac{P\cdot V}{n\cdot R_{u}}$

$T = \frac{(202.65\,kPa)\cdot(1.29\,m)^{3}}{(1\,kmole)\cdot(8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K} )}$

$T = 52.325\,K$

The thermal energy contained by the gas is:

$U = \frac{3}{2}\cdot (1\,kmole)\cdot (8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K})\cdot (52.325\,K)$

$U = 652.545\,kJ$

b) The physical model for the cat is constructed from Work-Energy Theorem:

$U = \frac{1}{2}\cdot m_{cat} \cdot v^{2}$

The speed of the cat is obtained by isolating the respective variable and the replacement of every known variable by numerical values:

$v = \sqrt{\frac{2 \cdot U}{m_{cat}}}$

$v = \sqrt{\frac{2\cdot (652.545 \times 10^{3}\,J)}{3\,kg} }$

$v \approx 659.568\,\frac{m}{s}$