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2. A gas that has a volume of 28 liters, a ternperature of 45 °C, and an unknown
pressure has its volume increased to 34 iters and its temperature decreased to
35 °C. If I measure the pressure after the change to be 2.0 atm, what was the
original pressure of the gas?

Respuesta :

ardni313

The  original pressure of the gas : 2.507 atm

Further explanation

Boyle's law and Gay Lussac's law  

[tex]\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}[/tex]

P1 = initial gas pressure (N/m² or Pa)  

V1 = initial gas volume (m³)  

P2 = final gas pressure  

V2 = final gas volume

T1 = initial gas temperature (K)  

T2 = final gas temperature  

V₁=28 L

T₁=45+273=318 K

P₂=2 atm

V₂=34 L

T₂=35+273=308 K

then initial pressure :

[tex]\tt P_1=\dfrac{P_2.V_2.T_1}{V_1.T_2}\\\\P_1=\dfrac{2\times 34\times 318}{28\times 308}\\\\P_1=2.507~atm[/tex]

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