in a car engine a 250 mL cylinder and piston system is filled with a mixture of 0.42g and 0.24 g of octane gas and oxygen gas, respectively. Both gases are at room temperature of 33 c. If the total gas mixture is compressed to a pressure of 3.7 atm before combustion, calculate the total volume of gas mixture at this point in time.

In this case, given the initial masses of both octane (C8H18, molar mass = 114 g/mol)) and oxygen (O2, molar mass = 32 g/mol), we can compute their moles and subsequently the total pressure via the ideal gas equation at 33 °C into the 250-mL:

[tex]n_{octane}=0.42g*\frac{1mol}{114g} =3.7x10^{-3}mol\\\\n_{oxygen}=0.24g*\frac{1mol}{32g} =7.5x10^{-3}mol\\\\n_T=7.5x10^{-3}mol+3.7x10^{-3}mol=0.0112mol\\\\P_1=\frac{n_TRT}{V_1}=\frac{0.0112mol*0.082\frac{atm*L}{mol*K}*(33+273.15)K}{0.25L}=1.1atm[/tex]

After that, via the Boyle's law, we compute the final pressure in the cylinder and piston system:

[tex]P_1V_1=P_2V_2\\\\V_2=\frac{P_1V_1}{P_2} =\frac{250mL*1.1atm}{3.7atm}\\ \\V_2=76.0mL[/tex]

Best regards!