Question

A gas sample enclosed in a rigid metal container at room temperature (20.0∘C) has an absolute pressure p1. The container is immersed in hot water until it warms to 40.0∘C. What is the new absolute pressure p2? Express your answer in terms of p1.

Answer 1

Answer : The new absolute pressure is, $1.068\times P_1$

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$

or,

$\frac{P_2}{P_1}=\frac{T_2}{T_1}$

where,

$P_1$ = initial pressure of gas

$P_2$ = final pressure of gas

$T_1$ = initial temperature of gas = $20.0^oC=273+20.0=293.0K$

$T_2$ = final temperature of gas = $40.0^oC=273+40.0=313.0K$

Now put all the given values in the above equation, we get:

$\frac{P_2}{P_1}=\frac{313.0K}{293.0K}$

$\frac{P_2}{P_1}=1.068$

$P_2=1.068\times P_1$

Therefore, the new absolute pressure is, $1.068\times P_1$

Answer 2

The new absolute pressure of the gas in terms of initial pressure is 1.068 P₁.

The given parameters;

• room temperature, T₁ = 20 ⁰C = 273 + 20 = 293 K
• initial pressure, P₁ = P₁
• final temperature, T₂ = 40 ⁰C = 40 + 273 = 313 K

The new absolute pressure of the gas at the given new temperature is determined by applying Pressure law;

$\frac{P_1}{P_2} = \frac{T_1}{T_2} \\\\P_2 = \frac{P_1T_2}{T_1} \\\\P_2 = \frac{P_1 \times 313}{293} \\\\P_2 = 1.068 \ P_1$

Thus, the new absolute pressure of the gas in terms of initial pressure is 1.068 P₁.

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