Compressing a spring increases its

If a gas has a volume of 1000 ML at a temperature of 23°C and a pressure of 100 mmhg, what is it’s volume under standard conditi

Colt1911 [192]

Answer:

119.7 mL.

Explanation:

From the general law of ideal gases:

PV = nRT.

where, P is the pressure of the gas.

V is the volume of the container.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas (K).

For the same no. of moles of the gas at two different (P, V, and T):

P₁V₁/T₁ = P₂V₂/T₂.

P₁ = 100.0 mmHg, V₁ = 1000.0 mL, T₁ = 23°C + 273 = 296 K.

P₂ = 1.0 atm = 760.0 mmHg (standard P), V₂ = ??? mL, T₂ = 0.0°C + 273 = 273.0 K (standard T).

∴ V₂ = (P₁V₁T₂)/(T₁P₂) = (100.0 mmHg)(1000.0 mL)(273.0 K)/(296 K)(760.0 mmHg) = 121.4 mL.

8 0

3 years ago

Calculate the final Celsius temperature when 634 L at 21 °C is compressed to 307 L.

Illusion [34]

Answer:

- 130.64°C.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have two different values of V and T:

V₁T₂ = V₂T₁

V₁ = 634.0 L, T₁ = 21.0°C + 273 = 294.0 K.

V₂ = 307.0 L, T₂ = ??? K.

∴ T₂ = V₂T₁/V₁ = (307.0 L)(294.0 K)/(634.0 L) = 142.36 K.

∴ T₂(°C) = 142.36 K - 273 = - 130.64°C.

5 0

3 years ago

Drag each tile to the correct location.

cestrela7 [59]

Carbon, helium, and sodium are monoatomic elements.

Hydrogen, iodine, and oxygen are diatomic elements.

3 0

3 years ago

Read 2 more answers

g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The heig

kumpel [21]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

$v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s$

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

$K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J$

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

$\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K$

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, $\Delta S_{env} = -1.18 J/K$

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

Learn more: brainly.com/question/22655760

6 0

2 years ago

Guys, look at the images. I'm so stuck, and I feel as if my brain is about to get ripped out.

Rama09 [41]

Answer:

You not alone lolI'm also tryna figure out the answer

3 0

2 years ago