A constant-volume gas thermometer is used to measure the temperature of an object. When the thermometer is in contact with water at its triple point (273.16K) the pressure in the thermometer is 8.500 × 104 Pa.

When it is in contact with the object the pressure is 9.650 × 104 Pa. The temperature of the object is:

Question

A constant-volume gas thermometer is used to measure the temperature of an object. When the thermometer is in contact with water at its triple point (273.16K) the pressure in the thermometer is 8.500 × 104 Pa.

When it is in contact with the object the pressure is 9.650 × 104 Pa. The temperature of the object is:

BlackTom AI · Accepted Answer

The problem describes a constant-volume gas thermometer, where pressure is proportional to temperature at fixed volume (P ∝ T).

First, identify the known point from the triple point of water: T1 = 273.16 K corresponds to P1 = 8.500 × 10^4 Pa.

Next, note the pressure when in contact with the object: P2 = 9.650 × 10^4 Pa. Since V is constant, the temperature T2 scales with pressure according to T2 = T1 × (P2 / P1).

Compute the pressure ratio: P2 / P1 = (9.650 × 10^4) / (8.500 × 10^4) = 9.650 / 8.500 ≈ 1.1353.

Now multiply by T1 to obtain T2: T2 ≈ 273.16 K × 1.1353 ≈ 310 K (approximately).

Therefore, the temperature of the object, as measured by the constant-volume gas thermometer, is about 310 Kelvin.

If you want the precise value, multiply 273.16 by 9.650/8.500 to get T2 ≈ 310.2 K, which rounds to 310 K in the given context.

PHYS_V 157 ALL SECTIONS 2024W1 Reading Quiz Week 2

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