A gas occupies $100.0 m L$ at $50^{o} C$ and 1 atm pressure. The gas is cooled at constant pressure so that volume is reduced to $50.0 m L$ . What is the final temperature of the gas?

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Solution

Given that:-
Initial volume $(V_{1}) = 100 m L$
Final volume $(V_{2}) = 50 m L$
Initial temperature $(T_{1}) = 50 ℃ = (273 + 50) K = 323 K$
Final temperature $(T_{2}) = ?$
From ideal gas law-
$P V = n R T$
Given that the pressure is constant, i.e.,
$V \propto T$
$\Rightarrow \frac{V_{1}}{V_{2}} = \frac{T_{1}}{T_{2}}$
$\Rightarrow \frac{100}{50} = \frac{323}{T_{2}}$
$\Rightarrow T_{2} = \frac{323 \times 50}{100} = 161.5 K$
Hence the requird answer is $161.5 K$ .

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A gas occupies 100.0mL at 50oC and 1 atm pressure. The gas is cooled at constant pressure so that volume is reduced to 50.0mL. What is the final temperature of the gas?

Given that:-Initial volume (V1)=100mLFinal volume (V2)=50mLInitial temperature (T1)=50℃=(273+50)K=323KFinal temperature (T2)=?From ideal gas law-PV=nRTGiven that the pressure is constant, i.e.,V∝T⇒V1V2=T1T2⇒10050=323T2⇒T2=323×50100=161.5KHence the requird answer is 161.5K.

A gas occupies $100.0 m L$ at $50^{o} C$ and 1 atm pressure. The gas is cooled at constant pressure so that volume is reduced to $50.0 m L$ . What is the final temperature of the gas?