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Question: At what temperature will the volume of a gas be twice the volume at \({27^ \circ }{\text{ }}C\) at a...

At what temperature will the volume of a gas be twice the volume at 27 C{27^ \circ }{\text{ }}C at a given pressure.
A. 327 C{327^ \circ }{\text{ }}C
B. 54 C{54^ \circ }{\text{ }}C
C. 127 C{127^ \circ }{\text{ }}C
D. 100 C{100^ \circ }{\text{ }}C

Explanation

Solution

In the given question we have to find the temperature of a gas when its temperature is changed. We have to use Charles’ Law to formulate an equation. Then by comparing both equations we will find our answer.

Complete step by step answer:
The pressure of the gas is constant. Let the initial volume of the gas be VV. According to the question the gas is at a temperature of 27 C{27^ \circ }{\text{ }}C.Converting degree Celsius scale into Kelvin scale we get,
K=C+273K = C + 273
where KK is in Kelvin and CC is in degree Celsius.
So, we get 27 C{27^ \circ }{\text{ }}C in Kelvin as (27+273)=300 K\left( {27 + 273} \right) = 300{\text{ }}K
According to Charles’ Law we get,
V=kTV = kT
where VV is the volume of the gas, TT is the temperature and kk is the constant of proportionality.

Now by substituting the values in Charles’ Law we get,
V=300kV = 300k
Arranging it we get,
V300=k(1)\dfrac{V}{{300}} = k - - - - - \left( 1 \right)
Now, when the volume is doubled means the final volume is then, 2V2V.
Let the temperature at which the volume becomes doble be TT.
Again, by substituting the values in Charles’ Law we get,
2V=Tk2V = Tk
Hence, arranging the equation we get,
2VT=k(2)\dfrac{{2V}}{T} = k - - - - - - \left( 2 \right)

Comparing equation (1)\left( 1 \right) and equation (2)\left( 2 \right) we get,
V300=2VT\dfrac{V}{{300}} = \dfrac{{2V}}{T}
By cross-multiplication we get,
T=600T = 600
Thus, the temperature is found to be 600 K600{\text{ }}K.
Now as the options are in degree-Celsius we have to convert the value from Kelvin to degree-Celsius. Converting Kelvin scale into degree-Celsius scale we get,
C=K273C = K - 273
where KK is in Kelvin and CC is in degree Celsius.
C=600273 C=327C = 600 - 273 \\\ \therefore C= 327
So, the final temperature is 327 C{327^ \circ }{\text{ }}C.

Hence the correct option is A.

Note: It must be noted that Charles’ Law is only applicable when the pressure of the given gas is constant. When the temperature of a gas is constant, we can follow Boyle’s Law and in case the volume is constant then we will use Gay-Lussac’s Law. It is very important that the mass of the gas in all cases must be constant also.