Question

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?

Answer 1

The original pressure of the gas : 2.507 atm

Further explanation

Boyle's law and Gay Lussac's law

`\tt (P_1.V_1)/(T_1)=(P_2.V_2)/(T_2)`

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 :

`\tt P_1=(P_2.V_2.T_1)/(V_1.T_2)\\\\P_1=(2* 34* 318)/(28* 308)\\\\P_1=2.507~atm`