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If the volume of a gas container at 32.0deg. C changes from 1.55 L to 755mL, what will hte final temp be?

I came up with...1.55/305=0.775/T = 152.5K which is equal to 425.5C. I sure I am wrong but can anyone help?

2007-05-15 04:33:16 · 8 answers · asked by Anonymous in Science & Mathematics Chemistry

8 answers

0°C = 273.15 K

So, 152.5 K = -120.65°C

2007-05-15 04:39:28 · answer #1 · answered by Dave_Stark 7 · 0 0

C = K - 273

= 152.5 - 273 = -120.5

for the other question, use this formula:

T1/V1 = T2/V2

solve for T2

T2 = T1V2/V1

for gas law calculations, temperatures must be in K. so you'll need to convert the given 32 deg C in to kelvin:

K = C + 273 = 32 + 273 = 305

you'll also need to get the given volumes into the same units. so you could convert the 1.55 L into mL:

1.55 L x [1000 mL / 1L] = 1550 mL

now you can plug your values into the equation:

T2 = T1V2/V1 = (305)(755)/1550 = 148.6 K

if you want that in celsius:

C = K - 273 = 148.6 - 273 = -124.4

2007-05-15 04:46:35 · answer #2 · answered by Anonymous · 0 0

If pressure is constant

we can use Charles' law

V1 / T1 = V2 / T2

32 °C = 273 + 32 = 305 K

1.55 / 305 = 0.775 / T2

0.00508 = 0.775 / T2

T2 = 152.5 K

152.5 K - 273 = - 120.5°C

2007-05-15 04:39:58 · answer #3 · answered by Anonymous · 0 0

0 degree C is approx. 273 K So, 75 degree C = 273 + 75 = 348 K

2016-05-18 21:28:31 · answer #4 · answered by ? 4 · 0 0

No, son, you're supposed to subtract 273 to get from K to C.

K= C +273
C= K-273

2007-05-15 04:38:42 · answer #5 · answered by Ellephatony 2 · 1 0

152.5deg K -273 = -120.5 deg C

2007-05-15 04:40:02 · answer #6 · answered by jcann17 5 · 0 0

152.5 kelvin = -120.65 degrees Celsius

2007-05-15 04:38:48 · answer #7 · answered by DanE 7 · 1 0

K = 273.15 + C

2007-05-15 06:02:06 · answer #8 · answered by shiara_blade 6 · 0 0

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