A gas is cooled from 365 K to 285 K while its volume changes from 12.8 L to 9.9 L. The initial pressure of the gas is 1.9 atm. The final pressure of the gas, rounded to the nearest tenth, is atm.
A gas is cooled from 365 K to 285 K while its volume changes from 12.8 L to 9.9 L. The initial pressure of the gas is 1.9 atm. The final pressure of the gas, rounded to the nearest tenth, is atm.
The correct answer to the question is : 1.9 atm.
EXPLANATION:
Let us consider the pressure, volume and temperature of a gas are denoted as P,V and T respectively.
Initial condition : Final condition:
\(P_{1} = 1.9\ atm\) \(P_{2} =\ ?\)
\(V_{1} =\ 12.8\ L\) \(V_{2} =\ 9.9\ L\)
\(T_{1}=\ 365\ K\) \(T_{2} =\ 285\ K\)
From combined equation of gas, we know that-
\(\frac{P_{1}V_{1}}{T_{1} } =\ \frac{P_{2}V_{2}}{T_{2}}\)
⇒ \(P_{2} =\ \frac{P_{1}V_{1}}{T_{1} } \times \frac{T_{2} }{V_{2} }\)
\(=\ \frac{1.9\times 12.8}{365}\times \frac{285}{9.9}\ atm\)
\(=\ 1.91814\ atm\)
\(=\ 1.9\ atm\) [ans]
Hence, the final pressure is 1.9 atm.