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Law Incorporation [45]
3 years ago
6

A cylinder containing an ideal gas has a volume of 2.6 m3 and a pressure of 1.5 × 105 Pa at a temperature of 300 K. The cylinder

is placed against a metal block that is maintained at 900 K and the gas expands as the pressure remains constant until the temperature of the gas reaches 900 K. The change in internal energy of the gas is +6.0 × 105 J. How much heat did the gas absorb?
Physics
1 answer:
frozen [14]3 years ago
5 0
<h2>Answer:13.5\times 10^{8} joules</h2>

Explanation:

From the first law of thermodynamics,

ΔQ=ΔU+W

Where Q is the heat given to the gas,

U is the internal energy of the gas,

W is the workdone by the gas.

When pressure is constant,

\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}

V_{2}=\frac{2.6\times 900}{300}=7.8m^{3}

When pressure is constant,W=PΔV

Where P is pressure and V is the volume of the gas.

Given P=1.5\times 10^{5}Pa

ΔV=7.8-2.6=5.2m^{3}

So,W=1.5\times 10^{5}\times 5.2=7.8\times 10^{5}J

Given that ΔU=6\times 10^{5}

So,ΔQ=6\times 10^{5}+7.8\times 10^{5}=13.8\times 10^{5}J

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A book with mass 2.3 kg sits on a table. What is the normal force on the
natita [175]

Explanation:

The table is level and there are no other forces on the book, so the normal force is equal to the weight.

N = mg

N = (2.3 kg) (9.8 m/s²)

N = 22.5 N

6 0
3 years ago
Read 2 more answers
A wheel starts from rest and has an angular acceleration that is given by α (t) = (6.0 rad/s4)t2. After it has turned through 10
marissa [1.9K]

Answer:

75 rad/s

Explanation:

The angular acceleration is the time rate of change of angular velocity. It is given by the formula:

α(t) = d/dt[ω(t)]

Hence: ω(t) = ∫a(t) dt

Also, angular velocity is the time rate of change of displacement. It is given by:

ω(t) = d/dt[θ(t)]

θ(t) = ∫w(t) dt

θ(t) = ∫∫α(t) dtdt

Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:

θ(t) = ∫∫α(t) dtdt

θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt

θ(t) = ∫[2t³]dt = t⁴/2 rad

θ(t) = t⁴/2 rad

At θ(t) = 10 rev = (10 *  2π) rad = 20π rad, we can find t:

20π = t⁴/2

40π = t⁴

t = ⁴√40π

t = 3.348 s

ω(t) = ∫α(t) dt = ∫6t² dt = 2t³

ω(t) = 2t³

ω(3.348) = 2(3.348)³ = 75 rad/s

7 0
3 years ago
An iron ball and an aluminum ball of mass 100 g each are heated to the same temperature and then cooled to a temperature of 20°
Y_Kistochka [10]
Becaused it the thing was cod that why
4 0
3 years ago
Help please! Will give brainliest to first correct answer!
Olegator [25]

Answer:

1) Addition of a catalyst

2) To change the reaction rate of slope B to look like slope A, simply add a catalyst to speed up the rate of reaction, giving you a higher amount of products in a shorter amount of time (line A)

Explanation:

1 and 2)Two things can alter the rate of a reaction, either the addition of a catylist which will not alter the composition of the products or reactants, but will accelerate the reaction time, or an increase in temperature will also increase the rate at which a reaction will occur.

You could choose temperature also and have the same result, it's your choice both are correct, but catalyst is the easiest.

   

8 0
3 years ago
The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization
ANEK [815]

Answer:

The vapor pressure at 60.6°C is 330.89 mmHg

Explanation:

Applying Clausius Clapeyron Equation

ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]

Where;

P₂ is the final vapor pressure of benzene = ?

P₁ is the initial vapor pressure of benzene = 40.1 mmHg

T₂ is the final temperature of benzene = 60.6°C = 333.6 K

T₁ is the initial temperature of benzene = 7.6°C = 280.6 K

ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol

R is gas rate = 8.314 J/mol.k

ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65  (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg

Therefore, the vapor pressure at 60.6°C is 330.89 mmHg

6 0
3 years ago
Read 2 more answers
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