1 Answers
Call
Talk Doctor Online in Bissoy App
A sample of a gas has a volume of of 2.0 liters at a pressure of 1.0 atmosphere. When the volume increases to 40 liters, at constant temperature, what will the pressure be?
This is an example of which states that the volume of a gas varies inversely with pressure, when mass and temperature are kept constant. Equation $V_1P_1=V_2P_2$, where $V$ is volume...
1 Answers 1 views
A gas occupies 3.30 liters and is at 850 mmHg. What would the pressure be in atm, if the new volume were increased to 5.63 liters?
This is a clear application of $"Boyle's Law"$ $Pprop1/V$ And thus, $P_1V_1=P_2V_2$ We note that $1*atm-=760*mm*Hg$ $P_2=(P_1V_1)/V_2$ $=$ $((850*mm*Hg)/(760*mm*Hg*atm^-1))xx(3.30*L)/(5.63*L)$ $=$ $??atm$ This is not a good question. It has...
1 Answers 1 views
A sample of gas occupies a volume of 50.0 millilters in a cylinder with a movable piston. The pressure of the sample is 0.00 atmosphere and the temperature is 200 K. What is the volume of the sample at STP?
How many atmospheres does the gas exert? You have it under no pressure, but it still occupies a volume. Eventually you will use the relationship: $(P_1V_1)/T_1=(P_2V_2)/T_2$, given a constant...
1 Answers 1 views
If I have 5.6 liters of gas in a piston at a pressure of 1.5 atm and compress the gas until its volume is 4.8 L, what will the new pressure inside the piston be?
At constant $T$, $P_1V_1=P_2V_2$....... We solve for $P_2=(P_1V_1)/V_2$, and clearly the $"RHS"$ will have the required units of pressure: $P_2=(1.5*atmxx5.6*L)/(4.8*L)=1.75*atm$ That $P_2>P_1$ is quite reasonable, because we are compressing the...
1 Answers 1 views
A sample of gas has a volume of 12 liters at 0°C and 380 torr. What will be its volume when the pressure is changed to 760 torr at a constant temperature?
$PV = nRT$ $P$ is pressure in atm (760 torr = 1 atm); $V$ is volume in liters; $n$ is the number of moles; $R$ is the constant 0.0821; $T$...
1 Answers 1 views
If the volume of the gas in a gas chamber is 500 mL at 277 K, what will be the volume of the gas in the chamber if the temperature is at 300 K?
Apply $"Volume"/"Temperature" = "Constant"$ at $"constant pressure"$ $V_1/T_1=V_2/T_2$ The initial volume is $V_1=500mL$ The initial temperature is $T_1=277K$ The final temperature is $T_2=300K$ The final volume is $V_2=T_2/T_1*V_1=300/277*500=541.5mL$
1 Answers 1 views
At a pressure 47 kPa, the gas in a cylinder has a volume of 12 liters. Assuming temperature remains the same, if the volume of the gas is decreased to 8 liters, what is the new pressure?
Boyles Law is the only inverse relationship between gas phase variable of the Empirical Gas Law set. That is ... For the Empirical the primary variables considered are Pressure...
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 occupies a volume of 50.0 milliliters in a cylinder with a movable piston. The pressure of the sample is 0.90 atmosphere and the temperature is 298 K. What is the volume of the sample at STP?
Use the $(P_1V_1)/T_1=(P_2V_2)/T_2$. $"STP=273.15 K and 1 atm"$ Given/Known $P_1="0.90 atm"$ $V_1="50.0 mL"$ $T_1="298 K"$ $P_2="1 atm"$ $T_2="273.15 K"$ Unknown $V_2$ Solution Rearrange the equation to isolate $V_2$ and solve....
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