A sample of krypton gas occupies a volume of 9.87 L at 51°C and 0.565 atm. If it is desired to decrease the volume of the gas sample to 8.05 L, while increasing its pressure to 0.678 atm, the temperature of the gas sample at the new volume and pressure must be °C.

44.1°C

Explanation:

Step 1:

Data obtained from the question include:

Initial volume (V1) = 9.87 L

Initial temperature (T1) = 51°C = 51°C + 273 = 324K

Initial pressure (P1) = 0.565 atm

Final volume (V2) = 8.05 L

Final pressure (P2) = 0.678 atm

Final temperature (T2) =..?

Step 2:

Determination of the new temperature of the gas.

The new temperature of the gas can be obtained by using the general gas equation as shown below:

P1V1 /T1 = P2V2 /T2

0.565 x 9.87/324 = 0.678 x 8.05/T2

Cross multiply

0.565 x 9.87 x T2 = 324 x 0.678 x 8.05

Divide both side by 0.565 x 9.87

T2 = (324 x 0.678 x 8.05)/(0.565 x 9.87)

T2 = 317.1K

Step 3:

Conversion of 317.1K to decree celsius.

This is illustrated below:

T (°C) = T(K) – 273

T (K) = 317.1K

T (°C) = 317.1 – 273

T (°C) = 44.1°C

Therefore, the new temperature of the gas is 44.1°C

a galvanic cell( aka voltaic cell) and electrolytic cell

explanation:

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