A 7.00 L sample of argon gas at 147 degres celcius exerts a pressure of 625 kPa. If the gas is compressed to 1.25 L and the temperature is lowered to 77 degres celcius, what will be its new pressure?

THE NEW PRESSURE OF THE GAS AT 77 °C AND 1.25 L IS 2919.7 kPa.

Explanation:

Using General Gas Equation;

P1 V1 / T1 = P2 V2 /T2

P1 = 625 kPa

V1 = 7 L

T1 = 147 = 147°C + 273 K = 420 K

V2 = 1.25 L

T2 = 77 °C = 273 + 77 K = 350 K

P2 = ?

Rearranging the equation by making P2 the subject of the formula. we have ;

P2 = P1 V1 T2 / V2 T1

Substitute the values into the equation;

P2 = 625 kPa * 7 L * 350 K / 1.25 L * 420 K

P2 = 1531250 *10^3 / 525

P2 = 2919.66 kPa.

The new pressure when the temperature of the gas is lower to 77 °C and compressed to 1.25 L is 2916.7 kPa.

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