Verified Answer :
Explanation:
The data is given as follows.
\(P_{1}\) = 1.12 atm, \(P_{2}\) = 1.67 atm
\(V_{1} = 15.7 m^{3}\), \(V_{2} = 11.2 m^{3}\)
\(T_{1} = 245 K\), \(T_{2}\) = ?
Therefore, calculate the final temperature using the relation as follows.
\(\frac{P_{1}V_{1}}{T_{1}}\) = \(\frac{P_{2}V_{2}}{T_{2}}\)
\(\frac{1.12 atm \times 15.7 m^{3}}{245}\) = \(\frac{1.67 atm \times 11.2 m^{3}}{T_{2}}\)
\(T_{2} = \frac{1.67 atm \times 11.2 m^{3} \times 245 K}{1.12 atm \times 15.7 m^{3}}\)
= \(\frac{4582.48 K}{17.584}\)
= 260.605 K
= 260.61 K (approx)
Thus, we can conclude that final temperature is 260.61 K.
The final temperature of the gas is 260.01 K.
The final temperature can be calculated by using the equation as follows:-
\(\frac{P_{1}V_{1} }{T_{1}} =\frac{P_{2}V_{2} }{T_{2}}\)........(1)
Here, \(P_1\) and \(P_2\) is the pressure of this gas before and after the changes.
\(V_1\) and \(V_2\) is volume of this gas before and after the changes.
\(T_1\) and \(T_2\) is the temperature (in degrees Kelvins) of this gas before and after the changes.
Given values:-
\(P_1=1.12\ atm, P_2=1.67\ atm\\\\V_1=15.7\ m^{3} , V_2=11.2\ m^{3}\\\\T_1=245\ K.T_2=?\)
Substitute all the values in the equation (1) as follows:-
\(\frac{1.12\ atm\times15.7\ m^{3}}{245\ K } =\frac{1.67\ atm\times11.2\ m^{3}}{T_2 }\\T_2=\frac{1.67\ atm\times11.2\ m^{3}\times245\K}{1.12\ atm\times15.7\ m^{3} }\\=260.60\ K\)
Hence, the final temperature of the gas is 260.60 K.
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