A rigid, sealed container that can hold 26 L of gas is filled to a pressure of 5.97 atm at 374 °C. The pressure suddenly decreases to 3.64 atm. What is
the new temperature inside the container, in units of °C?

121 °C

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 5.97 atm

Initial temperature (T₁) = 374 °C

Final pressure (P₂) = 3.64 atm

Final temperature (T₂) =?

NOTE: Volume = constant

Next, we shall convert 374 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 374 °C

Initial temperature (T₁) = 374 °C + 273

Initial temperature (T₁) = 647 K

Next, we shall determine the final temperature. This can be obtained as follow:

Initial pressure (P₁) = 5.97 atm

Initial temperature (T₁) = 647 K

Final pressure (P₂) = 3.64 atm

Final temperature (T₂) =?

P₁ / T₁ = P₂ / T₂

5.97 / 647 = 3.64 / T₂

Cross multiply

5.97 × T₂ = 647 × 3.64

5.97 × T₂ = 2355.08

Divide both side by 5.97

T₂ = 2355.08 / 5.97

T₂ = 394 K

Finally, we shall convert 394 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

Final temperature (T₂) = 394 K

Final temperature (T₂) = 394 – 273

Final temperature (T₂) = 121 °C

Thus, the new temperature is 121 °C

2b2t

explanation:

2b2t

there are two significant figures.