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Question: An open flask contains air at 27\({}^\circ C\). Calculate the temperature at which it should be heat...

An open flask contains air at 27C{}^\circ C. Calculate the temperature at which it should be heated so that 1/3rd of the air measured at 27C{}^\circ C escapes out.

Explanation

Solution

Air at 27C{}^\circ C behaves like an ideal gas. Convert the unit of temperature from celsius to Kelvin by adding 273k to the given temperature. Use the formula for ideal gas, here volume and pressure are constant.

Complete step by step solution:
From your chemistry lessons you have learned about the ideal gas and its equation. In the question the pressure and volume are kept constant. So there will be relation between moles and temperature.This can also be seen through ideal gas equation,
Ideal gas equation, PV=nRT
So, the relation between temperature and moles at constant pressure and volume is,

n1T1=n2T2{{n}_{1}}{{T}_{1}}={{n}_{2}}{{T}_{2}}…………(1)
Where, n= moles
T = Temperature
In the question initial temperature is given (T1)({{T}_{1}}) = 27C{}^\circ C
But we have to convert the temperature into Kelvin,
So, T1{{T}_{1}} = 27+ 273 K = 300K
Let us take initial no. of moles n1{{n}_{1}} at 300k = n
Now, Let the new temperature T2{{T}_{2}} be= T K
At temperature T K the no. of moles will be = 13n\dfrac{1}{3}n
So, the no. of moles (n2)({{n}_{2}}) left at T K = n13n=23nn-\dfrac{1}{3}n=\dfrac{2}{3}n
Now , put all the values in formula (1),

n1T1=n2T2{{n}_{1}}{{T}_{1}}={{n}_{2}}{{T}_{2}}

n×300=23n×Tn\times 300=\dfrac{2}{3}n\times T
T = 450 K
T = 450-273 = 177C{}^\circ C.

Note: In any question if temperature is given in celsius firstly convert it into Kelvin and then solve the question and at last if the answer is in celsius then again convert the answer in Kelvin to Celsius by subtraction 273 from the temperature. Final mole will be the left mole at T K and not the 1/3 n because it is the escape amount not the left one.