Question

A gas sample enclosed in a rigid metal container at room temperature (20.0∘C∘C) has an absolute pressure p1p1p_1. The container is immersed in hot water until it warms to 40.0∘C∘C. What is the new absolute pressure p2p2p_2

Answer 1

Answer:

${P_{2}} =1.0682P_{1}$

Explanation:

This problem is based on the application of ideal gas equation. Ideal gas equation is very important tool in thermodynamics to solve many problems. So, here we solve this problem using ideal gas equation.

The temperature when the pressure is $P_{1}$ be $T_{1}$

We convert temperature in SI unit,

 $T_{1}$= $20^{0}C= 20 +273.15K=293.15K$

The temperature when pressure is $P_{2}$  be $T_{2}$

$T_{2}=40^{0}C= 40 +273.15K=313.15K$

By ideal gas equation,

$\frac{P_{1} }{P_{2}} =\frac{T_{1} }{T_{2}}$

$\frac{P_{1} }{P_{2}} =\frac{293.15}{313.15}$

$\frac{P_{2} }{P_{1}} =\frac{313.15}{293.15}=1.0682$

so,

${P_{2}} =1.0682P_{1}$

Pressure become 1.0682 times.