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Romashka-Z-Leto [24]
3 years ago
13

Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h

eat during the process and the change of entropy of the gas. The isobaric heat capacity of the gas is Cp.m 28.253 J-K1-mol1.
Chemistry
1 answer:
bija089 [108]3 years ago
7 0

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

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What is the approximate partial pressure of oxygen at 1500 m?
elena55 [62]

The approximate partial pressure of oxygen at 1500 m is 0.18 atm.

<h3>Pressure of the air</h3>

From barometric pressure table;

1500 m = 85 kPa

1 kPa = 0.0099 atm

85 kPa = ?

= 0.84 atm

Oxygen percentage in air  = 21%

<h3>Partial pressure of oxygen in air</h3>

P = 0.21 x P(total)

P = 0.21 x 0.84 atm

P = 0.18 atm

Thus, the approximate partial pressure of oxygen at 1500 m is 0.18 atm.

Learn more about partial pressure here: brainly.com/question/19813237

#SPJ11

3 0
2 years ago
What is the [OH-]if the pH of a solution 10.62
Lerok [7]
As we know,

                     pH + pOH  =  14
Or,
                     pOH  =  14 - pH

Putting Value of pH,

                    pOH  =  14 - 10.62

                    pOH  =  3.38

pOH is converted into [OH⁻] as follow,

                   [OH⁻]  =  10⁻pOH             ∴ -pOH is in power, 10 means antilog
                                                               press shift and click log button
                   [OH⁻]  =  10⁻³°³⁸

                   [OH⁻]  =  0.0004168

                   [OH⁻]  =  4.16 × 10⁻⁴
5 0
3 years ago
Explain the scientific word meaning of random
astraxan [27]
Odd, unusual, or unexpected.
5 0
3 years ago
The atmospheric pressure on the surface of Venus is 6.84X10^4. Calculate the atmospheric pressure in atm and torr. Round each of
quester [9]

Answer:

0.675 atm

513 Torr

Explanation:

Given is that, the atmospheric pressure on the surface of Venus is

6.84 X 10⁴ Pa.

1 atm (atmospheric pressure) is equal to 101325 pascal (Pa).

To convert divide the pressure value by 101325.

Pressure in atm = \frac{6.84 \times 10^{4} }{101325}

= 0.675055 atm

Rounding it off to 3 significant digits: 0.675 atm

Now,  one Torr is 133.322 Pa. For conversion, divide the pressure value by 133.322.

Pressure in Torr = \frac{6.84 \times 10^{4} }{133.322}

=513.04219 Torr

Rounding it off to 3 significant digits: 513 Torr

5 0
2 years ago
9. A tank truck carries 34 000 L of sulfuric acid. The density
creativ13 [48]

Answer:

<h2>62,560 kg</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

mass = Density × volume

From the question

volume = 34,000 L

density = 1.84 kg/L

We have

mass = 1.84 × 34,000 = 62,560

We have the final answer as

<h3>62,560 kg</h3>

Hope this helps you

7 0
2 years ago
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