# 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?

Feb 10, 2017

$\text{412.6 K}$

#### Explanation:

$\text{Let us recall the gas law}$

$\text{PV = nRT}$

Our question is that a quantity of a gas at a temperature of
223K has a volume of ${\text{100.0 dm}}^{3}$.

To what temperature must the gas be raised, while the pressure is kept constant, to give a volume of ${\text{185 dm}}^{3}$ ?

So let;s rearrange the formula

$P = \frac{\text{(nRT)}}{V}$

$W h e r e V = 100.0 {\mathrm{dm}}^{3}$

$\text{Where p = is unknown or x}$

$\text{Where R = gas constant law which is 0.0821L}$

$\text{Where n = n is number of moles of gas}$

So let us plug in the variables

${\text{100 dm}}^{3}$ should be converted in litres first.

$\text{1 dm"^3 = "1 L}$

So

$\text{100 dm"^3 = "100 L}$

$\frac{n \cdot 0.0821 L \cdot 223 K}{100 L} = \text{18.309n"/"100L} = 0.18309 n$

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

The expression would be

$185 {\mathrm{dm}}^{3} = \frac{n \cdot 0.0821 L \cdot T}{0.18309 n}$

Cut out the both n

$185 {\mathrm{dm}}^{3} = 0.0821 \cdot \frac{T}{\text{0.18309}}$

$\text{so T = Let's solve your equation step-by-step.}$

$185 = \frac{\text{0.0821t}}{0.18309}$

Step 1: Multiply both sides by 0.18309.

$185 = \frac{\text{0.0821t}}{0.18309}$

$\left(185\right) \cdot \left(0.18309\right) = \left(\text{0.0821t} \cdot 0.18309\right) \cdot \left(0.18309\right) = 33.87165 = 0.0821 t$

Step 2: Flip the equation.

$0.0821 t = 33.87165$

Step 3: Divide both sides by 0.0821.

$\frac{\text{0.0821t}}{0.0821} = \frac{33.87165}{0.0821}$

$t = 412.565773$

$T = \text{412.6 K}$