Solving a word problem .(kelvins & degrees Celsius)?

By how many kelvins would you have to lower the temperature of a sample of gas to change its volume from 1.75 L to 50.0 mL if the initial temperature was 375 K? Express the answer both in kelvins and in degrees Celsius.

Answer:
Do what the other answerers did by using the correct "Ideal Gas
Law."

PV = nRT

Assuming constant P and n, then ...

V1 / T1 = V2 / T2

V1 = 1.75 L
V2 = 0.05 L
T1 = 375 K
T2 = ? (solve, then convert to C)

Recall that the ideal gas law is PV=nRT. If we take the ratio of the two conditions, call them states 1 and 2, we would have:

P1*V1/(P2*V2) = n1*R*T1/(n2*R*T2)

Since pressure, P, and number of moles, n, are not changing and R is a constant, they cancel to give:
V1/V2 = T1/T2

So, T2 = T1(V2/V1)
T2 = 375 * (0.05/1.75) = 10.7 K

So, you would have to reduce the temperature by 375-10.7 = 364.3 degrees C / K (since change in temperature is the same in C and K)

EDIT - corrected my answer since apparently I can't do simple subtraction in my head. :)

Charle's Law. Direct proportion at constant Pressure.
Volume Decrease <===> Temperature Decrease.
V1 x T2 = V2 x T1
1.75 L x T2 = 0.050 L x 375K (102°C)
T2 = (0.050 x 375) ÷ 1.75
Final temp. (T2) = 10.7 K = -262°C.
A reduction of 364° in both C & K

(Check by proportion...
0.050 ÷ 1.75 = 0.29
0.29 x 375K = 10.7 K)

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Solving a word problem .(kelvins & degrees Celsius)?

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