Chapter 5: Q5.29P (page 228)

If $1.47 \times 10^{- 3}$ of argon occupies a 75.0-mL container at $26 . ° C$ , what is the pressure (in torr)?

Short Answer

Expert verified

Answer

The pressure of the gas is 365 torr.

Step by step solution

Ideal gas law

The ideal gas law equation is,

$P V = n R T$

The ideal gas law for the given condition can be written as,

$P = \frac{nRT}{V}$

Here,

V is the volume ( $75.0 m L = 0.075 L$ ).

T is the temperature ( $26 ° C = 299 K$ ).

n is the number of moles ( $1.47 \times 10^{- 3} m o l$ ).

Determination of the pressure.

The pressure of the gas is,

$P = \frac{n R T}{V} P = \frac{(1.47 \times 10^{- 3} m o l) \times 0.0821 L a t K^{- 1} m o l^{- 1} \times 299 K}{0.075 L} P = 0.48 a t m$

Conversion of the unit of pressure from atm to torr,

$1 t o r r = 0.00132 a t m$

And,

$0.481 a t m \times \frac{1 torr}{0.00132}$

Thus, the pressure of the gas is 365 torr.

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If $1.47 \times 10^{- 3}$ of argon occupies a 75.0-mL container at $26 . ° C$ , what is the pressure (in torr)?

Short Answer

Step by step solution

Ideal gas law

Determination of the pressure.

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If 1.47×10-3of argon occupies a 75.0-mL container at 26.°C, what is the pressure (in torr)?

Short Answer

Step by step solution

Ideal gas law

Determination of the pressure.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Kinetics

Making Measurements

Nuclear Chemistry

Organic Chemistry

Chemistry Branches

Physical Chemistry

Study anywhere. Anytime. Across all devices.

If $1.47 \times 10^{- 3}$ of argon occupies a 75.0-mL container at $26 . ° C$ , what is the pressure (in torr)?