A 3.50-L gas sample at 20°C and a pressure of 86.7 kPa expands to a volume of 8.00 L. The final pressure of the gas is 56.7 kPa. What is the final temperature of the gas, in degrees Celsius?

$(P_1V_1)/T_1=(P_2V_2)/T_2$, 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 $"absolute temperature."$ $P_2=(1.00*atmxx250*mLxx313*K)/(500*mLxx293*K)~=1/2*atm.$

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...

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

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

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$T_"Ar"=223+273=500$...

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...

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...

You know that the temperature of the gas is being held constant, so you should be able to identify that you're dealing with a situation covered by here....

This isthe an example of which states the volume of a gas that is held at a constant pressure varies directly with the Kelvin temperature. The formula used to solve...

$(P_1V_1)/T_1=(P_2V_2)/T_2$ You need to convert degC to Kelvin by adding 273: $:.(200xx25)/298=(250xxV_2)/273$ $:.V_2=(200xx25xx273)/(298xx250)$ $:.V_2=18.32"L"$