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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....$V_2=(P_1V_1)/T_1xxT_2/P_2$ $=(150*kPaxx200*L)/(273*K)xx(546*K)/(600*kPa)=??*L$
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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, $V_2=(V_1xxT_2)/T_1$ $=$ $(4.0*m^3xx400*cancelK)/(200*cancelK)$ $=$ $??*m^3$.
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Under a pressure of 200 kPa, a confined gas has a volume of 2,500 cubic meters. The pressure acting on the gas is increased to 500 kPa. What is the new volume of the gas if the temperature remains the same?
Use the : $P_1 V_1 = P_2 V_2$ . substituting and making the units consistent: $(200xx10^3)(2500)=(500xx10^3)(V_2)$. solve for $V_2$ .
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A gas takes up a volume of 19.5 liters, has a pressure of 2.9 atm, and a temperature of 5.5°C. If I raise the temperature to 65°C and lower the pressure to 1.5 atm, what is the new volume of the gas?
From the $PV=nRT$ we can conclude that $n=(PV)/(RT)$. If the pressure $P$, the volume $V$ and the temperature $T$ of the gas change between two points, this change can be...
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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$T_"Ar"=223+273=500$...
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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...
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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.................. $V_2=(P_1xxV_1xxT_2)/(T_1xxP_2)$ $=(2.31*atmxx17.5*Lxx350*K)/(299*Kxx1.75*atm)<=30*L.$
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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...
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The pressure of 250 kPa acting on 15 cubic meters of gas at a temperature of 100 K is increased to 500 kPa and the volume is increased to 30 cubic meters. What is the temperature of the gas at this new pressure and volume?
This is an example of the . The equation to use is $(P_1V_1)/(T_1)=(P_2V_2)/(T_2)$. Given Initial pressure, $P_1="250 kPa"$ Initial volume, $V_1="15 m"^3"$ Initial temperature, $T_1="100 K"$ Final pressure, $P_2="500 kPa"$...
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Fifty liters of gas is kept at a temperature of 200 K and under pressure of 15 atm. The temperature of the gas is increased to 400 K. The pressure is decreased to 7.5 atm. What is the resulting volume of the gas?
Even without doing any calculations, you should be able to look at the information given and predict that the volume of the gas will increase after temperature is increased and...
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