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A sample of methane that initially occupies a 850.0 mL at 500.0 kPa and 500.0K is compressed to a volume of 7000.0mL. To what temperature will a gas need to be cooled to lower the pressure of gas to 200.0 kPa
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@Somy COuld you possibly help real quick
sure \[\sf \frac{ P_1V_1 }{ T_1 } = \frac{ P_2V_2 }{ T_2 }\] \(\sf P_1\) = 500.0 kPa \(\sf \color{red}{convert ~ to ~ Pa}\) \(\sf V_1\) = 850.0 mL \(\sf \color{red}{convert~ to ~m^3}\) \(\sf T_1\) = 500.0K \(\sf P_2\) = 200.0 kPa \(\sf \color{red}{convert ~ to ~ Pa}\) \(\sf V_2\) = 7000.0mL \(\sf \color{red}{convert~ to ~m^3}\) \(\sf T_2\) = ? After converting values to proper units Put them into formula and get the ans
Okay thanks
No problem :)
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