If the Sun were the size of a small exercise ball (about 0.5 meters (m) in

A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1

balandron [24]

Answer:

The answer is " $143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}$ "

Explanation:

For point a:

Energy balance equation:

$\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\$

$W=0\\\\Q=0\\\\m_e=0$

From the above equation:

$\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\$

because the rate of air entering the tank that is $h_i$ constant.

$\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\$

Since the tank was initially empty and the inlet is constant hence, $m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\$

Interpolate the enthalpy between $T = 300 \ K \ and\ T=295\ K$ . The surrounding air

temperature:

$T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}$

Substituting the value from ideal gas:

$\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}$

Follow the ideal gas table.

The $u_2= 298.33\ \frac{kJ}{kg}$ and between temperature $T =410 \ K \ and\ T=240\ K.$

Interpolate

$\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}$

Substitute values from the table.

$\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\$

For point b:

Consider the ideal gas equation. therefore, p is pressure, V is the volume, m is mass of gas. $\bar{R} \ is\ \frac{R}{M}$ (M is the molar mass of the gas that is $28.97 \ \frac{kg}{mol}$ and R is gas constant), and T is the temperature.

$n=\frac{pV}{TR}\\\\$

$=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\$

For point c:

Entropy is given by the following formula:

$\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}$

5 0

3 years ago

According to Newton's first law of motion, what would happen to a car traveling at a constant velocity of 55 MPH to the west in

mart [117]

Explanation:

Newton's first law of motion says that an object will be at rest or it will be in motion until an unbalanced force acts on it.
In this case, a car traveling at a constant velocity of 55 mph to the west in the absence of an unbalanced force.

It would mean that the car will continue to move with a constant velocity of 55 mph to the west due to Newton's first law of motion.

6 0

3 years ago

Classify pH values of 9.1, 1.2, and 5.7 as basic, acidic, or very acidic.

kramer

9.1 basic since > 7
1.2 VERY acidic << 7
<span>5.7 acidic < 7 </span>

4 0

3 years ago

The strike zone is an area over home plate extending approximately from

Masteriza [31]

Answer:

if it is a true or false, the answer is true

Explanation:

3 0

2 years ago

Which statement best describes the movement of electrical current when a capacitor is used in a circuit?

lukranit [14]

Answer:

The discharge of a capacitor changes the direction of the current

Explanation:

Capacitor is used to store charges. Changing and discharging a capacitor changes the direction of the current.

When a capacitor is charging, current flows towards the positive plate and away from the negative plate. When the capacitor is discharging, current flows away from the positive and flows towards the negative plate, in the opposite direction. During these two processes (charging and discharging), it can be seen that the direction of the current is changed.

Therefore, the discharge of a capacitor changes the direction of the current

5 0

3 years ago