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3 years ago

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The volume of a gas is directly related to the temperature of the gas. at 30 degrees celsius a certain gas has a volume of 125 c

ubic centimeters. what will the new volume be at 50 degrees celsius?

2 answers:

pentagon [3]3 years ago

8 0

<u>Answer:</u> The new volume is $133.25cm^3$

<u>Explanation:</u>

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

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=125cm^3\\T_1=30^oC=(30+273)K=303K\\V_2=?cm^3\\T_2=50^oC=(50+273)K=323K$

Putting values in above equation, we get:

$\frac{125cm^3}{303K}=\frac{V_2}{323K}\\\\V_2=\frac{125\times 323}{303}=133.25cm^3$

Hence, the new volume is $133.25cm^3$

tigry1 [53]3 years ago

5 0

To solve this we use <span>Charles' law whitch states that V1/T1=V2/T2.
So if T1=30 C, V1=123 cm^3 and T2=50 C -> V2=(V1*T2)/T1
T2=208.33 cm^3

</span>

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Answer:

Part A)

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Part B)

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Explanation:

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Let's use the parabolic motion equation to solve it. Let's define the variables:

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For the second stone

$0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}$

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If we solve the equation (1) we will have:

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We can do the same procedure for the equation (2)

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We can use the equation of the horizontal position here.

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$x_{f1}=x_{i1}+vcos(30)t_{1}$

$x_{f1}=0+13.86*t_{1}$

$x_{f1}=13.86*t_{1}$

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$x_{1}=0+13.86*t_{1}$

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Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

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We can solve this problem by using the first law of thermodynamics, which states that the change in internal energy of a system is given by the equation:

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Therefore, in this problem, we have

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