Question

Caed for this question.

Answer 1

Answer:

0.962 atm.

97.4 kPa.

731 torr.

14.1 psi.

97,434.6 Pa.

Explanation:

Hello.

In this case, given the available factors equaling 1 atm of pressure, each required pressure turns out:

- Atmospheres: 1 atm = 760 mmHg:

$p=731mmHg*\frac{1atm}{760mmHg} =0.962atm$

- Kilopascals:: 101.3 kPa = 760 mmHg:

$p=731mmHg*\frac{101.3kPa}{760mmHg} =97.4kPa$

- Torrs: 760 torr = 760 mmHg:

$p=731mmHg*\frac{760 torr}{760mmHg} =731 torr$

- Pounds per square inch: 14.69 psi = 760 mmHg:

$p=731mmHg*\frac{14.69}{760mmHg} =14.1psi$

- Pascals: 101300 Pa = 760 mmHg:

$p=731mmHg*\frac{101300Pa}{760mmHg} \\\\p=97,434.6Pa$

Best regards.