Convert 532 torr to kPa.

Convert 532 torr to kPa. Remember that multiple steps may be required in the conversion. STARTING AMOUNT ADD FACTOR ANSWER RESET 01325 105 1000 0.709 1.01325 0.700 101.325 0.001 105 14.70 760 98692 70.9 532 pSI torr bar Pa mm Hg atm kPa

The Correct Answer and Explanation is:

To convert 532 torr to kilopascals (kPa), we can use a two-step conversion process:

Step 1: Understand the conversion factors

The key conversion factor between torr and pascals (Pa) is:

  • 1 torr = 133.322 Pa

Also, remember:

  • 1,000 Pa = 1 kPa

Step 2: Convert 532 torr to pascals (Pa)

532 torr×133.322 Patorr=70927.3 Pa532\ \text{torr} \times 133.322\ \frac{\text{Pa}}{\text{torr}} = 70927.3\ \text{Pa}532 torr×133.322 torrPa​=70927.3 Pa

Step 3: Convert pascals to kilopascals

70927.3 Pa÷1000=70.93 kPa70927.3\ \text{Pa} \div 1000 = 70.93\ \text{kPa}70927.3 Pa÷1000=70.93 kPa


Final Answer: 532 torr = 70.93 kPa


🧠 Explanation

When performing pressure conversions, it’s important to use the correct units and conversion factors. In this case, we are converting from torr, a unit of pressure originally defined as 1 millimeter of mercury (mm Hg), to kilopascals (kPa), a unit in the metric system. Because the two units are not directly interchangeable, a two-step conversion is needed.

The first step involves converting from torr to pascals. This works because the pressure unit torr is defined as 133.322 pascals, based on the precise physical measurement of one millimeter of mercury. Multiplying the starting pressure in torr by this factor converts it to pascals.

The second step is to convert from pascals to kilopascals. Since “kilo” means one thousand, 1,000 pascals equal 1 kilopascal. Dividing the pascal value by 1,000 gives the answer in kilopascals.

This step-by-step process ensures accuracy when changing between different pressure measurement systems. Understanding how to use and apply conversion factors is critical in scientific and engineering work, especially in chemistry and physics, where pressure units appear in gas laws and fluid dynamics. Doing this conversion correctly shows both unit awareness and proper dimensional analysis, which is a core principle in all science disciplines.

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