A sample of helium gas occupies 14.7 L at 23degree C and 0.956 atm. What volume

Question

A sample of helium gas occupies 14.7 L at 23degree C and 0.956 atm. What volume will it occupy at 40degree C and 1.20 atm? V_1 = 12.4L V_2 = 7 T_1 = 23degree C + +273 = 296K T_2 = 40degree C + 273 = 313K P_1 = 0.956 atm P_2 = 1.20 atm 19.5 L 20.4 L 11.1 L 12.4 L 14.9 L The pressure of 5.4 L of nitrogen gas in a flexible container is decreased to one-half its original pressure, and its absolute temperature is increased to double the original temperature. The volume is now 2.7 L. 5.4 L. 11 L. 22 L. 1.4 L.

Explanation / Answer

3.

Given that

P1 =0.956 atm

V1= 14.7 L

T1= 23 C or 296 K

And T 2= 40 C= 313 K

P2 = 1.20 atm

We know that

P1V1/T1 = P2V2/T2

V2 = P1V1/T1* T2/P2

= 0.956 *14.7/296 * 313/ 1.20

= 12.4 L

4.

P1 = P1, V1 = 5.4 L, T1 = T1

P2 = 1/2*P1, V2 = ?, T2 = 2*T1

Therefore;

P1*V1/T1 = P2*V2/T2

P1*5.4/T1 = 0.5*P1*V2/(2*T1)

solve for V2

5.4= 0.5*V2/(2)

V2 = 5.4*2/0.5 = 21.6 L

Or 22 L

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