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

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

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Answer 1 · 2017-02-10T14:56:44

$412.6 K$ $"412.6 K"$

Explanation:

$Let us recall the gas law$ $"Let us recall the gas law"$

$PV = nRT$ $"PV = nRT"$

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

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

So let;s rearrange the formula

$P = \frac{(nRT)}{V}$ $P = "(nRT)"/V$

$W h e r e V = 100.0 d m^{3}$ $Where V = 100.0dm^3$

$Where p = is unknown or x$ $"Where p = is unknown or x"$

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

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

So let us plug in the variables

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

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

So

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

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

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

The expression would be

$185 d m^{3} = \frac{n \cdot 0.0821 L \cdot T}{0.18309 n}$ $185dm^3 = (n * 0.0821L * T)/(0.18309n)$

Cut out the both n

$185 d m^{3} = 0.0821 \cdot \frac{T}{0.18309}$ $185dm^3 = 0.0821 * T/"0.18309"$

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

$185 = \frac{0.0821t}{0.18309}$ $185="0.0821t"/0.18309$

Step 1: Multiply both sides by 0.18309.

$185 = \frac{0.0821t}{0.18309}$ $185="0.0821t"/0.18309$

$(185) \cdot (0.18309) = (0.0821t \cdot 0.18309) \cdot (0.18309) = 33.87165 = 0.0821 t$ $(185) * (0.18309) = ("0.0821t" * 0.18309) * (0.18309) = 33.87165=0.0821t$

Step 2: Flip the equation.

$0.0821 t = 33.87165$ $0.0821t=33.87165$

Step 3: Divide both sides by 0.0821.

$\frac{0.0821t}{0.0821} = \frac{33.87165}{0.0821}$ $"0.0821t"/0.0821 = 33.87165/0.0821$

$t = 412.565773$ $t=412.565773$

Answer:

$T = 412.6 K$ $T= "412.6 K"$

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

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