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3 years ago

A gas sample occupies 4.2 L at a pressure of 101 kPa. What volume will it occupy if the pressure is increased to 235 kPa?

2 answers:

7 0

For this we want to use Boyle's Law. Boyle’s law states that the pressure and volume of a fixed quantity of a gas are inversely proportional under constant temperature conditions. The formula for this is P1V1 = P2V2. We want to solve this out so it equals V2 (Volume 2). So P1V1 / P2 = V2. Then plug in your values for the variables. So (101)(4.2) / 235 = V2; so 424.2 / 235 = V2. The final volume equals 1.81. I hope this helps, If not I am very sorry.

grin007 [14]3 years ago

3 0

0.9L ....... may also be the answer youre looking for

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irga5000 [103]

The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

The given parameters;

Pressure on the 13 th floor, P₁ = 35 PSI
Distance between each floor, d = 10 ft

The vertical pressure of the water is calculated as follows;

$P = \rho gh\\\\\frac{P}{h} = \rho g\\\\\frac{P}{h} = k\\\\\frac{P_1}{h_1} = \frac{P_2}{h_2} \\\\$

The vertical height of the first floor from the 13th floor = 130 ft

The vertical height of the 13 ft floor = 10 ft

$P_1 = \frac{P_2 h_1}{h_2} \\\\P_1 = \frac{35 \times 130}{10} \\\\P_1 = 455 \ PSI$

Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

Learn more about vertical height and pressure here: brainly.com/question/15691554

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3 years ago

Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const

kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is L= 20cm = 0.2m

Thermal conductivity of the wall is K = 2.79 W/m·K

Temperature at the left side surface is T₁ = 50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

$Q_{conduction} = Q_{convection}$

Expression for the heat conduction process is

$Q_{conduction} = \frac{K(T_1 - T)}{L}$

Expression for the heat convection process is

$Q_{convection} = h(T_2 - T)$

Substitute the expressions of conduction and convection in equation above

$Q_{conduction} = Q_{convection}$

$\frac{K(T_1 - T_2)}{L} = h(T_2 - T)$

Substitute the values in above equation

$\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC$

Now heat flux through the wall can be calculated as

$q_{flux} = Q_{conduction} \\\\q_{flux} = \frac{K(T_1 - T_2)}{L}\\\\q_{flux} = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2$

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

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