The volume of a sample of a gas at 273 Kelvin and is 100.0 L. If the volume is decreased to 50.0 L at constant pressure, what will be the new temperature of the gas?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

This is an example of , which states that the volume of a given amount of a gas is directly proportional to its Kelvin temperature. The formula to use for this law is $V_1/T_1=V_2/T_2$.

Given
$V_1="100.0 L"$
$T_1="273 K"$
$V_2="50.0 L"$

Unknown
$T_2$

Solution
Rearrange the formula to isolate $T_2$. Substitute the given values into the formula and solve.

$V_1/T_1=V_2/T_2$

$T_2=(T_1V_2)/V_1$

$T_2=(273"K" xx 50.0cancel"L")/(100.0cancel"L")$

$T_2="136.5 K"="137 K"$ rounded to three

The proportionality between volume and temperature can be easily seen in this problem. As the volume decreased by half, the Kelvin temperature decreased by half.

A very good tutorial on the can be found at .

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
You have a sample of neon gas at a certain pressure, volume, and temperature. You double the volume, double the number of moles of neon, and double the Kelvin temperature. How does the final pressure (Pf) compare to the original pressure (Po) ?

The final pressure will double as well. If you use the equation, $PV = nRT$, to express the initial and the final states of your gas sample, you can...

1 Answers 1 views
A 20.0 mL sample of a gas is at 546 K and has a pressure of 6.0 atm. If the temperature is changed to 273 K and the pressure to 2.0 atm, the new volume of the gas is what?

$V_2=(P_1xxV_1xxT_2)/(P_2xxT_1)$ We immediately see that we will get an answer in $"terms of volume"$, so we're batting on a firm wicket. $V_2=(6.0*atmxx20.0*mLxx273*K)/(2.0*atmxx546*K)=??*mL$

1 Answers 1 views
A gas takes up a volume of 17.5 liters, has a pressure of 2.31 atm, and a temperature of 299 K. lf the temperature is increased to 350 K and the pressure is decreased to 1.75 atm, what is the new volume of the gas?

And so.................. $V_2=(P_1xxV_1xxT_2)/(T_1xxP_2)$ $=(2.31*atmxx17.5*Lxx350*K)/(299*Kxx1.75*atm)<=30*L.$

1 Answers 1 views
A gas occupies a volume of 444 mL at 273 K and 79.0 kPa. What is the final kelvin temperature when the volume of the gas is change 880 mL and the pressure is changed to 38.7 kPa?

Using the you can solve for your missing variable. $(P1V1)/(T1)=(P2V2)/(T2)$ P1 = 79.0 kPa V1 = 444 mL T1 = 273 K P2 = 38.7 kPa V2 = 880...

1 Answers 1 views
The volume of a sample of a gas at 273°C is 200.0 L. If the volume is decreased to 100.0 L at constant pressure, what will be the new temperature of the gas?

states that when pressure is held constant, the temperature and volume of a gas are directly proportional, so that if one goes up, so does the other. Note: The...

1 Answers 1 views
The pressure of a sample of dry air is held constant at 2.25 atm while the temperature is decreased rom 100 °C to 7 °C. The original volume of the sample is 43 L. What is the final volume of the sample?

We use old .......$VpropT$ at constant pressure. $(V_1)/T_1=(V_2)/T_2$ And solve for $V_2=(V_1)/(T_1)xxT_2$ $(43*L)/(373*K)xx280*K=??L$

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 helium gas has a volume of 900 milliliters and a pressure of 2.50 atm at 298 K. What is the new pressure when the temperature is changed to 336 K and the volume is decreased to 450 milliliters?

So following the gas law: $P_2 = (P_1V_1)/T_1 xx T_2/V_2 = (2.50*atmxx900xx10^(-3)cancelLxx336cancelK )/(298cancelKxx450xx10^(-3)cancelL)$ $=$ $>5*atm ?$ Pressure change is reasonable because we compressing the gas, AND raising the temperature.

1 Answers 1 views
A 3.00 liter sample of gas is at 288 K and 1.00 atm. If the pressure of the gas is increased to 2.00 atm and its volume is decreased to 1.50 liters, what will be the Kelvin temperature of the sample?

When you are given this much information in the context of a generic, unnamed gas, it's a good idea to consider the : $PV = nRT$ Before we...

1 Answers 1 views
A $73*mL$ volume of gas enclosed in a piston at a pressure of $100*kPa$, and a temperature of $0$ $""^@C$ was heated to a temperature of $80$ $""^@C$ and the new volume occupied was $4530*mL$. What is the new pressure that the gas exerts?

Temperature must be quoted on the $"Kelvin scale"$, and so... $P_2=(P_1V_1)/T_1xxT_2/V_2$, which will certainly give us an answer with the required units of pressure. Why so? And thus........... $P_2=(100*kPaxx73.0*mL)/(273.15*K)xx(353.15*K)/(4530*mL)$ $~=2*kPa$...

1 Answers 1 views
X