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3 years ago

A typical refrigerator is kept at 4˚C, and a soda can has a pressure of

Chemistry

1 answer:

ss7ja [257]3 years ago

7 0

Answer: The new pressure will be 1.42 atm

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=1.18atm\\T_1=4^0C=(4+273)K=277K\\P_2=?\\T_2=60^0C=(60+273)K=333K$

Putting values in above equation, we get:

$\frac{1.18}{277}=\frac{P_2}{333}\\\\P_2=1.42$

Hence, the new pressure will be 1.42 atm

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Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress

ElenaW [278]

Answer:

(a) $W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) $W_{polytropic}=7.579\frac{kJ}{mol}$

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant $k$ should be computed for air as an ideal gas by:

$\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\$

$0.2856=1-\frac{1}{k}\\\\k=1.4$

Next, we compute the final temperature:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K$

Thus, the work is computed by:

$W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) In this case, since $n$ is given, we compute the final temperature as well:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K$

And the isentropic work:

$W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}$

$W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}$

Regards.

8 0

3 years ago

Please help me out its due tmr :/

lesantik [10]

Answer:

using this for points, did you get the answer because I can try and help?

Explanation:

7 0

3 years ago

The most common cooling mechanism for cloud formation is ________.

KatRina [158]

The answer is "Rising and expanding air or atmosphere cooling".
To form a cloud the air must be cooled to the temperature of dew point. when there is expansion of air, it cools to the dew point and thus the formation of cloud happens. This process is very common and used for the formation of clouds.

8 0

3 years ago

A mixture of helium and hydrogen gases, at a total pressure of 751 mm Hg, contains 0.447 grams of helium and 0.193 grams of hydr

Margaret [11]

Answer:

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Explanation:

4 0

3 years ago

Complete the balanced neutralization equation

lbvjy [14]

Answer:

2 HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + 2 H₂O

Explanation:

Let's consider the reaction between acetic acid and strontium hydroxide. This is a neutralization reaction, in which an acid reacts with a base to form salt and water. The unbalanced equation is:

HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + H₂O

We have 1 acetate ion to the left and 2 to the right, so we will multiply HC₂H₃O₂(aq) by 2.

2 HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + H₂O

Finally, we multiply water by 2 to get the balanced equation.

2 HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + 2 H₂O

8 0

3 years ago