A cylinder containing 5.20 L of NO gas at 1.20 atm and 32 o C is compressed under 6.00 atm pressure to a volume of 2.08 L. What is the final temperature of the gas?

337 °C

Explanation:

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

Initial volume (V1) = 5.20 L

Initial pressure (P1) = 1.20 atm

Initial temperature (T1) = 32 °C

Final pressure (P2) = 6 atm

Final volume (V2) = 2.08 L

Final temperature (T2) =?

Next, we shall convert 32 °C to Kelvin temperature. This can be obtained as follow:

T (K) = T (°C) + 273

Initial temperature (T1) = 32 °C

Initial temperature (T1) = 32 °C + 273

Initial temperature (T1) = 307 K

Next, we shall determine the final temperature of the gas as illustrated below:

Initial volume (V1) = 5.20 L

Initial pressure (P1) = 1.20 atm

Initial temperature (T1) = 305 K

Final pressure (P2) = 6 atm

Final volume (V2) = 2.08 L

Final temperature (T2) =?

P1V1 /T1 = P2V2 /T2

1.2 × 5.2 / 305 = 6 × 2.08 / T2

Cross multiply

1.2 × 5.2 × T2 = 305 × 6 × 2.08

6.24 × T2 = 3806.7

Divide both side by 6.24

T2 = 3806.7 / 6.24

T2 = 610 K

Finally, we shall convert 610 K to celsius temperature. This can be obtained as follow:

T (°C) = T(K) – 273

T(K) = 610 K

T (°C) = 610 – 273

T (°C) = 337 °C

Thus, the final temperature of the gas is 337 °C.

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