425 mL of neon gas at - 12°C and 788 mmHg

expands into a 2.40 L container while the

pressure decreases to 577 mmHg. What is the

temperature of the gas in °C?

806.23 °C

Explanation:

The question requires we calculate the new temperature of the gas in °C.

We are given;

Initial volume of neon gas, V₁ = 425 mL or 0.425 L Initial temperature of neon gas, T₁ = - 12°C

When doing questions on gas law, we always use temperature in Kelvin.

To convert temperature from °C to Kelvin we always use;

K = °C + 273

Therefore, Initial temperature, T₁ = 261 K

Initial Pressure, P₁ = 788 mmHg New pressure of the gas, P₂ = 577 mmHg New volume of the gas, V₂ = 2.40 L

To calculate the new temperature, we are going to use the combined gas equation.

According to the combined gas law, P₁V₁/T₁ = P₂V₂/T₂ Rearranging the equation;

T₂ = P₂V₂T₁ / P₁V₁

Therefore;

T₂ = (577 mmHg * 2.4 L * 261 K) : (788 mmHg * 0.425 L)

= 1079.226 K

= 1079.23 K

But, °C = K - 273

= 1079.23 K - 273

Therefore, T₂ = 806.23 °C

Therefore, the new temperature is 806.23°C