At what temperature would the volume of a gas be equal to 0.796 L, if the gas had a volume of 1180 mL at 417 K? (round to 3 sig figs)

281 K

Explanation:

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

Initial volume (V₁) = 1180 mL

Initial temperature (T₁) = 417 K

Final volume (V₂) = 0.796 L

Final temperature (T₂) =?

Next, we shall convert 0.796 L to millilitres (mL). This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.796 L = 0.796 L × 1000 mL / 1 L

0.796 L = 796 mL

Finally, we shall determine the final temperature as follow:

Initial volume (V₁) = 1180 mL

Initial temperature (T₁) = 417 K

Final volume (V₂) = 796 mL

Final temperature (T₂) =?

V₁/T₁ = V₂/T₂

1180 / 417 = 796/T₂

Cross multiply

1180 × T₂ = 417 × 796

1180 × T₂ = 331932

Divide both side by 1180

T₂ = 331932 / 1180

T₂ = 281 K

Therefore, the final temperature of the of the gas is 281 K

more info ? ? i don't understand what the question is

explanation:

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.