A sample of sulfur hexafluoride gas occupies 8.62 L at 221 degrees Celsius. Assuming that the pressure remains constant, what temperature (in Celsius) is needed to reduce the volume to 2.61 L?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

Your variables, $"L"$ and $""^@"C"$ represent volume and temperature. The gas law that involves the relationship between the volume of a gas and its temperature is .

Charles' law states that the volume of a given amount of gas held at constant pressure is directly proportional to its Kelvin temperature. This means that as the temperature increases, the volume also increases, and vice-versa. The equation used to solve gas problems involving Charles' law is:

$V_1/T_1=V_2/T_2$,

where the temperatures must be in Kelvins. Add $273.15$ to the Celsius temperature to get the Kelvin temperature.

Organize your data:

Known

$V_1="8.62 L"$

$T_1="221"^@"C" + 273.15="494 K"$

$V_2="2.61 L"$

Unknown

$T_2=?$

Solution

Rearrange the equation given above to isolate $"T_2$. Insert your data into the new equation and solve.

$T_2=(V_2T_1)/V_1$

$T_2=(2.61color(red)cancel(color(black)("L"))xx494"K")/(8.62color(red)cancel(color(black)("L")))="150. K"$

To convert from Kelvins to degrees Celsius, subtract $273.15$ from the Kelvin temperature.

$"150. K"-273.15"="-123^@"C"$

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
Elemental sulfur occurs as octatomie molecules, $S_8$. What mass of fluorine gas is needed to react completely with 28.6 g of sulfur to form sulfar hexafluoride?

We need (i) a stoichiometric equation: $S(s) + 3F_2(g) rarr SF_6(g)$. And (ii) the molar equivalence of sulfur and fluorine $=$ $(28.6*g)/(32.06*g*mol^-1)$ $=$ $0.892*mol$. And thus $0.892xx3xx38.00*g*mol^-1$ $~=$ $100*g$...

1 Answers 1 views
Elemental sulfur occurs as octatomic molecules, $S_8$. What mass (g) of fluorine gas is needed to react completely with 17.8 grams of sulfur to form sulfur hexafluoride?

Start with a balanced equation. $"S"_8+"24F"_2$$rarr$$"8SF"_6"$ Then determine the molar masses of $"S"_8"$ and $"F"_2"$ by multiplying the subscript of each element by its from in g/mol. $"S"_8":$ $(8xx32.06 "g/mol")="256.48...

1 Answers 1 views
A250. mL sample of gas at 1.00 atm and 20 degrees Celsius has the temperature increased to 40 degrees Celsius and the volume increased to 500. mL. What is the new pressure?

$(P_1V_1)/T_1=(P_2V_2)/T_2$, given a constant quantity of gas. And thus, $(P_1xxV_1xxT_2)/(V_2xxT_1)=P_2$, even given this format, we can see the problem is dimensionally sound. We do need to use $"absolute temperature."$ $P_2=(1.00*atmxx250*mLxx313*K)/(500*mLxx293*K)~=1/2*atm.$

1 Answers 1 views
Two hundred liters of gas at zero degrees Celsius are kept under a pressure of 150 kPa. The temperature of the gas is raised to 273 degrees Celsius and the pressure is increased to 600 kPa. What is the new volume?

...according to the combined gas equation... We solve for....$V_2=(P_1V_1)/T_1xxT_2/P_2$ $=(150*kPaxx200*L)/(273*K)xx(546*K)/(600*kPa)=??*L$

1 Answers 1 views
The volume of 4.0 cubic meters of gas is kept under constant pressure. Its temperature is increased from a minus 73 degrees Celsius to 127 degrees Celsius. What is the new volume of the gas?

Since $V$ $prop$ $T$, $V$ $=$ $kT$. Alternatively, (if we solve for $k$), $k=V_1/T_1=V_2/T_2$. So, $V_2=(V_1xxT_2)/T_1$ $=$ $(4.0*m^3xx400*cancelK)/(200*cancelK)$ $=$ $??*m^3$.

1 Answers 1 views
After 0.600 L of Ar at 1.29 atm and 223 degrees Celsius is mixed with 0.200 L of O2 at 535 torr and 106 degrees Celsius in a 400.-mL flask at 23 degrees Celsius, what is the pressure in the flask?

Concept of is to be applied here. Let the partial pressure of $Ar$ in the final mixture be $p_"Ar"$ Initially before mixing $ Ar$ had Pressure $P_"Ar"=1.29$ atm Temperature$T_"Ar"=223+273=500$...

1 Answers 1 views
After 0.600 L of Ar at 1.47 atm and 221 degrees Celsius is mixed with 0.200 L of O2 at 441 torr and 115 degrees Celsius in a 400.-mL flask at 24 degrees Celsius, what is the pressure in the flask?

Here's how I go about doing this. What we can do is use the ideal gas equation for both $"Ar"$ and $"O"_2$ to find the total moles. Then we...

1 Answers 1 views
A sample of nitrogen collected in the laboratory occupies a volume of 725 ml, at standard pressure. What volume will the gas occupy at a pressure of 49.3 kPa, assuming the temperature remains constant?

Here I assume it is $101.3*kPa-=1*atm$... We use old ...$PV=k$...and so $P_1V_1=P_2V_2$, given CONSTANT temperature, and a GIVEN quantity of gas. And thus $V_2=(P_1V_1)/P_2-=(725*mLxx101.3*kPa)/(49.3*kPa)$ ...so $V_2~=1.5*L$ ...that is the volume...

1 Answers 1 views
A sample of carbon dioxide gas at a pressure of 1.07 atm and a temperature of 166 °C, occupies a volume of 686 mL. If the gas is heated at constant pressure until its volume is 913 mL, the temperature of the gas sample will be ? °C.

The idea here is that the volume and the temperature of a gas have a direct relationship when the pressure and the number of moles of gas are being kept...

1 Answers 1 views
A sample of gas at 35degree Celsius and 1 atm occupies a volume of 37.5L. At what temp should the gas be kept, if it required to reduce the volume to 3.0 litre at the same pressure?

Use Charles Law. Temperature and Volume are directly related using degrees Kelvin. ( if the pressure and number of molecules are kept constant. $ V_1/T_1 = V_2/T_2$ $...

1 Answers 1 views
X