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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 $"absolute temperature."$ $P_2=(1.00*atmxx250*mLxx313*K)/(500*mLxx293*K)~=1/2*atm.$
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The temperature of a fixed mass of gas in a rigid container is raised from 28.00 degrees Celsius to 88.00 degrees Celsius. The initial pressure was 1000 mmHg. What is the new pressure, after heating?
This is an example of Gay Lussac's temperature-pressure gas law , which states that the pressure of a gas held at constant volume, is directly proportional to the temperature in...
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The pressure acting on 60 cubic meters of gas is raised from 236 kPa to 354 kPa. The temperature is kept constant. What new volume does the gas occupy?
states that the volume of an enclosed gas is inversely proportional tot he pressure exerted on it, provided the temperature remains constant. $therefore P_1V_1=P_2V_2$ $therefore V_2=(P_1V_1)/P_2$ $=(236xx60)/354$ $=40m^3$
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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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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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Given a gas under $600*"mm Hg"$ pressure, and at $303.15*K$ temperature, that occupies a volume of $2.02*L$, what volume will it occupy at $273.15*K$ temperature, under $750*"mm Hg"$ pressure?
And $1*"bar"-=(1*"bar")/(1.01325*atm*"bar"^-1)=0.987*atm$ I do all this rigmarole BECAUSE I know that $1*atm$ will support a column of mercury that is $760*mm$ high........... And hence in terms of the length of...
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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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