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2 years ago

A sample of neon gas in a bulb is at 149.05 °C and 349.84 kPa. If the pressure drops

Chemistry

1 answer:

KonstantinChe [14]2 years ago

6 0

The new temperature (in °C) of the gas, given the data is –148.20 °C

<h3>Data obtained from the question </h3>

Initial temperature (T₁) = 149.05 °C = 149.05 + 273 = 422.05 K
Initial pressure (P₁) = 349.84 KPa
Volume = constant
New pressure (P₂) = 103.45 KPa
New temperature (T₂) =?

<h3>How to determine the new temperature </h3>

The new temperature of the gas can be obtained by using the combined gas equation as illustrated below:

P₁V₁ / T₁ = P₂V₂ / T₂

Since the volume is constant, we have:

P₁ / T₁ = P₂ / T₂

349.84 / 422.05 = 103.45 / T₂

Cross multiply

349.84 × T₂ = 103.45 × 422.05

Divide both side by 349.84

T₂ = (103.45 × 422.05) / 349.84

T₂ = 124.80 K

Subtract 273 from 124.80 K to express in degree celsius

T₂ = 124.80 – 273

T₂ = –148.20 °C

Learn more about gas laws:

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