Question

An engineer turns the temperature of a gas down, holding the volume constant at 1,250 L.Initially, the temperature of the gas was 732.°C and had a pressure of 294. kPa. Determine the pressure of the gas when the temperature decreased to 156°C.

Answer 1

Answer:

125.49kPa

Explanations:

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

$\begin{gathered} P\alpha T \\ P=kT \\ T=\frac{P_1}{T_1}=\frac{P_2}{T_2} \end{gathered}$

Given the following parameters

Iinitial pressure P1 = 294kPa

Initial temperature = 732.°C = 732+273 = 1005K

Final temperature = 156°C + 273 = 429K

Required

New pressure P2

Substitute the given parameters

$\begin{gathered} P_2=\frac{P_1T_2}{T_1} \\ P_2=\frac{294\times429}{1005} \\ P_2=125.49kPa \end{gathered}$

Therefore the pressure of the gas when the temperature decreased to 156°C is 125.49kPa