To what temperature a gas at 227°C must be cooled such that its volume is reduced to 1/5 of the initial volume pressure remaining constant?

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Solution

Step 1- Write down the given terms:
T₁=227K T₂ = ?
Initial Volume V₁= V
Final Volume V₂= $\frac{V}{5}$
Step 2- Write the formula:
$\frac{V_{1}}{T_{1}}$ = $\frac{V_{2}}{T_{2}}$
Step 3- Put the given values
$\frac{V}{227}$ = $\frac{V}{5}$ x $\frac{1}{T_{2}}$
T₂= $\frac{227}{5}$ = 45.4 °C
or T₂= 45.4+ 273.15
= 318.55 K.
Step 4: Result
Hence, The temperature is 45.4 °C or 318.55 K.

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To what temperature a gas at 227°C must be cooled such that its volume is reduced to 1/5 of the initial volume pressure remaining constant?

Step 1- Write down the given terms:T1 =227K T2 = ?Initial Volume V1= VFinal Volume V2= V5Step 2- Write the formula:V1T1 = V2T2Step 3- Put the given valuesV227 = V5 x 1T2T2= 2275 = 45.4 °Cor T2= 45.4+ 273.15= 318.55 K.Step 4: ResultHence, The temperature is 45.4 °C or 318.55 K.