A quantity of a gas at a temperature of #223# #K# has a volume of #100.0# #dm^3# To what temperature must the gas be raised, while the pressure is kept constant, to give a volume of #185# #dm^3#?

1 Answer

Answer:

#"412.6 K"#

Explanation:

#"Let us recall the gas law"#

#"PV = nRT"#

Our question is that a quantity of a gas at a temperature of
223K has a volume of #"100.0 dm"^3#.

To what temperature must the gas be raised, while the pressure is kept constant, to give a volume of #"185 dm"^3# ?

So let;s rearrange the formula

#P = "(nRT)"/V#

#Where V = 100.0dm^3#

#"Where p = is unknown or x"#

#"Where R = gas constant law which is 0.0821L"#

#"Where n = n is number of moles of gas"#

So let us plug in the variables

#"100 dm"^3# should be converted in litres first.

#"1 dm"^3 = "1 L"#

So

#"100 dm"^3 = "100 L"#

# (n * 0.0821L * 223K) /(100L) = "18.309n"/"100L" = 0.18309n #

So if pressure is constant to get a 185dm^3

The expression would be

#185dm^3 = (n * 0.0821L * T)/(0.18309n)#

Cut out the both n

#185dm^3 = 0.0821 * T/"0.18309"#

#"so T = Let's solve your equation step-by-step."#

#185="0.0821t"/0.18309#

Step 1: Multiply both sides by 0.18309.

#185="0.0821t"/0.18309#

#(185) * (0.18309) = ("0.0821t" * 0.18309) * (0.18309) = 33.87165=0.0821t#

Step 2: Flip the equation.

#0.0821t=33.87165#

Step 3: Divide both sides by 0.0821.

#"0.0821t"/0.0821 = 33.87165/0.0821#

#t=412.565773#

Answer:

#T= "412.6 K"#