Question

a rigid container holds a gas at a pressure of 55kPa and a tempature of -100 C. what will the pressure be when the tempature is increased to 200 C

Answer 1

See answer below

Step-by-step explanation:

Hi there,

This question is employing the Gas Laws, specifically the relationship between internal gaseous pressure and temperature, when volume and gas quantity are constant; Gay-Lussac's Law:

`(P)/(T) =k` or
`(P_1)/(T_1) =(P_2)/(T_2)`

However, we will need to convert the pressures and temperatures into standard units. Pressure must be put in terms of (atm) and temperature in Kelvin (K).

`P_1=55kPa*(1 \ atm)/(101.3 \ kPa) = 0.543 \ atm\\T_1 = -100 \ C + 273.15 = 173.15 \ K\\T_2 = 200 \ C + 273.15 = 473.15 \ K`

Now, solve for
`P_2` as this is what was asked for:

`P_2 = T_2(P_1)/(T_1)= (473.15K)*(0.543 \ atm)/(173.15K) = 1.484 \ atm`

If needed, convert back into kPa, so 1.484 * 101.3 kPa = 150.3 kPa.

thanks,