Question

Part A A gas sample enclosed in a rigid metal container at room temperature (20.0∘C) has...

Part A

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.

p2 =

1.0682p1

Part B

Nitrogen gas is introduced into a large deflated plastic bag. No gas is allowed to escape, but as more and more nitrogen is added, the bag inflates to accommodate it. The pressure of the gas within the bag remains at 1.00 atm and its temperature remains at room temperature(20.0∘C). How many moles n have been introduced into the bag by the time its volume reaches 22.4 L?

Express your answer in moles.

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Answer #1
Concepts and reason

The concepts required to solve the given question is ideal gas equation.

Initially, find the relation between the absolute pressure and the new absolute pressure Р.
. Later, calculate the number of moles introduced into the bag by using the ideal gas equation.

Fundamentals

The expression for the ideal gas equation is as follows:

PV nRT

Here, P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

The expression for the relation between the pressure and the temperature is as follows:

Р«т

Here, P is the pressure and T is the temperature, and signifies that the pressure is directly proportional to the temperature.

(A)

Write the equation Р«т
for the pressure and the new absolute pressure for the pressure Р.
using the equation Р«т
as follows:

Р Т
Р. Т,

Solve for Р.
.

PT,
2

Now, substitute 40.0 °C
for T2
and 20 °C
for in the equation PT,
2
.

Р(40 °C)
(20 °C)
P(40 273 K)
(20+273 K)
1.0682P

(B)

Rearrange the equation PV nRT
as follows:

PV
п-
RT

Substitute 101.325 Pa
for P, 22.4 m
3
for V, 8.314 J/mol K
for R, and 20.0 C
for T in the equation PV
п-
RT
.

(101.325 Pa) (22.4 m)
(8.314 J/mol-K) (20.0 °C)
(101.325 Pa)(22.4 m)
(8.314 J/mol -K)(20.0+273 K)
=0.932 moles
Il

Ans: Part A

The new absolute pressure Р.
in terms of is equal to 1.0682P
.

Part B

The number of moles introduced into the bag is equal to 0.932 moles.

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