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You are given an ammonia gas with an initial pressure of 20°C and an initial pressure of 2.5atm and a final pressure of 760mmHg. You are required to find the final temperature of your ammonia gas at 760mmHg. The equation to be used in this problem is the Gay-Lussac’s law which has a formula of P1V1/T1=P2V2/T2. Let’s say that your P1=2.5atm, T1=20°C and P2=760mmHg. However, you will notice that your V1&2 is unknown, therefore, in this problem the volume of your ammonia gas before and after confinement is constant. Then your Gay-Lussac’s equation becomes P1/T1=P2/V2. Before you plug in all the values of your variables, you must see to it that the units of your values are the same. For pressure, 1atm=760mmHg. Now that your pressure units are the same, input the values to your equation. 2.5atm/20°C=1atm/T2 T2=1atm/(2.5atm)( 20°C) T2=9°C
I think the answer is 8 coz 1/2.5 =0.4 X 20 = 8