Question

If the temperature of a gas is raised from 30°C to 60°C, what happens to the pressure?

Answer 1

Answer:

The pressure increases by 10% of the original pressure

Thus the new pressure is 1.1 times the original pressure.

Explanation:

We are given;

• Initial temperature as 30°C, but K = °C + 273.15
• Thus, Initial temperature, T1 =303.15 K
• Final temperature, T2 is 333.15 K

We are required to state what happens to the pressure;

• We are going to base our arguments to Pressure law;
• According to pressure law, the pressure of a gas and its temperature are directly proportional at a constant volume

• That is; P α T
• Therefore, at varying pressure and temperature

$\frac{P1}{T1}=\frac{P2}{T2}$

Assuming the initial pressure, P1 is P

Rearranging the formula;

[tex]P2=\frac{P1T2}{T1}[/tex]

$P2=\frac{(P)(333.15K)}{303.15K}$

     $P2 = 1.099P$

                 = 1.10 P

The new pressure becomes 1.10P

This means the pressure has increased by 10%

We can conclude that, the new pressure will be 1.1 times the original pressure.