Question

I have pressure of 65 kPa and starting temperature of 25°C. If I raise the temperature to 167°C, what is the new pressure? (Equation: K= 273 +°CO 90 KPaO 100 kPaO 96 kPaO 97 kPa

Answer 1

96kPa. Option C is correct

Explanations:

According to Gay Lussac's law, the pressure of a given mass of gas is directly proportional to the temperature provided that the volume is constant. Mathematically:

`\begin{gathered} P\alpha T \\ P=kT \\ k=(P_1)/(T_1)=(P_2)/(T_2) \end{gathered}`

Substituting the given parameters to determine the new pressure P2

`\begin{gathered} P_2=(P_1T_2)/(T_1) \\ P_2=(65*(167+273))/((25+273)) \\ P_2=(65*440)/(298) \\ P_2=(28600)/(298)=95.97kPa \\ P_2\approx96kPa \end{gathered}`

Therefore the new pressure is 96kPa