Chemistry, 05.10.2019 09:01 zhellyyyyy

The volume of a gas-filled balloon is 30.0 l at 40 °c and 153 kpa pressure. what volume will the balloon have at standard temperature and pressure (273.15 k and 101.3 kpa)?
a. 17.3 l
b. 23.7 l
c. 39.5 l
d. 51.9 l

a gas that has a volume of 28 liters, a temperature of 45 °c, and an unknown pressure, has its volume increased to 34 liters and its temperature decreased to 35 °c. if i measure the pressure after the change to be 2.0 atm, what was the original pressure of the gas?
a. 1.5 atm
b. 1.7 atm
c. 2.8 atm
d. 2.5 atm

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Answer from: alexandria3498

Hello!

The volume of a gas-filled balloon is 30.0 L at 40 °C and 153 kPa pressure. What volume will the balloon have at standard temperature and pressure (273.15 K and 101.3 kPa)?

a. 17.3 L

b. 23.7 L

c. 39.5 L

d. 51.9 L

We have the following data:

V1 (initial volume) = 30 L

T1 (initial temperature) = 40ºC (in Kelvin)

TK = TºC + 273.15

TK = 40 + 273.15 → T1 (initial temperature) = 313.15 K

P1 (initial pressure) = 153 kPa

V2 (final volume) = ? (in L)

T2 (final temperature) = 273.15 K

P2 (final pressure) = 101.3 kPa

Now, we apply the data of the variables above to the General Equation of Gases, let's see:

$\dfrac{P_1*V_1}{T_1} =\dfrac{P_2*V_2}{T_2}$

$\dfrac{153*30}{313.15} =\dfrac{101.3*V_2}{273.15}$

$\dfrac{4590}{313.15} =\dfrac{101.3\:V_2}{273.15}$

multiply the means by the extremes

$313.15*101.3\:V_2 = 4590*273.15$

$31722.095\:V_2 = 1253758.5$

$V_2 = \dfrac{1253758.5}{31722.095}$

$\boxed{\boxed{V_2 \approx 39.5\:L}}\:\:\:\:\:\:\bf\blue{\checkmark}$

c. 39.5 L

_______________________

A gas that has a volume of 28 liters, a temperature of 45 °C, and an unknown pressure, has its volume increased to 34 liters and its temperature decreased to 35 °C. If I measure the pressure after the change to be 2.0 atm, what was the original pressure of the gas?

a. 1.5 atm

b. 1.7 atm

c. 2.8 atm

d. 2.5 atm

We have the following data:

V1 (initial volume) = 28 L

T1 (initial temperature) = 45ºC (in Kelvin)

TK = TºC + 273.15

TK = 45 + 273.15 → T1 (initial temperature) = 318.15 K

P1 (initial pressure) = ? (in atm)

V2 (final volume) = 34 L

T2 (final temperature) = 35ºC (in Kelvin)

TK = TºC + 273.15

TK = 35 + 273.15 → T2 (final temperature) = 308.15 K

P2 (final pressure) = 2 atm

Now, we apply the data of the variables above to the General Equation of Gases, let's see:

$\dfrac{P_1*V_1}{T_1} =\dfrac{P_2*V_2}{T_2}$

$\dfrac{P_1*28}{318.15} =\dfrac{2*34}{308.15}$

$\dfrac{28\:P_1}{318.15} =\dfrac{68}{308.15}$

multiply the means by the extremes

$28\:P_1*308.15 = 318.15*68$

$8628.2\:P_2 = 21634.2$

$P_2 = \dfrac{21634.2}{8628.2}$

$P_2 = 2.507382768... \to \boxed{\boxed{P_2 \approx 2.5\:atm}}\:\:\:\:\:\:\bf\blue{\checkmark}$

d. 2.5 atm

_______________________

$\bf\green{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

Answer from: Quest

c. is the answer

by love gaming

Answer from: Quest

the formula for the conversion of temperature from degree celsius to kelvin is:

$k = ^{o}c + 273.15$

the sample of water at $38^{o}c$ is converted to kelvin temperature by using the above formula as:

$k = 38^{o}c + 273.15 = 311.15 k$

on comparing the two samples of water at $38^{o}c$ or we can say that at $311.15 k$ and at $295 k$ respectively. the difference between the temperatures of two samples of water is:

$311.15 k - 295 k = 16.15 k$

hence, one can say that the sample of water at $38^{o}c$ is greater than the sample of water at $295 k$ by $16.15 k$ .

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The volume of a gas-filled balloon is 30.0 l at 40 °c and 153 kpa pressure. what volume will the bal...

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