Question

use the references to access important values if needed for this question.
(a) a sample of krypton gas at a pressure of 0.537 atm and a temperature of 208 °c, occupies a volume of 477 ml. if the gas is cooled at constant pressure until its volume is 374 ml, the temperature of the gas sample will be
°c.
(b) a sample of carbon dioxide gas at a pressure of 904 mm hg and a temperature of 24 °c, occupies a volume of 11.6 liters. if the gas is heated at constant pressure to a temperature of 70 °c, the volume of the gas sample will be
l.
submit answer
try another version
3 item attempts remaining

Explanation:

Step1: Convert temperatures to Kelvin

For part (A):
Initial temperature \(T_1 = 208^{\circ}C+273 = 481\ K\)
Final temperature \(T_2\) (let's find it).
Initial volume \(V_1 = 477\ mL\)
Final volume \(V_2 = 374\ mL\)
Using Charles's law \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\), so \(T_2=\frac{V_2T_1}{V_1}\)
\(T_2=\frac{374\times481}{477}\ K\approx377\ K\)
\(T_2 = 377 - 273=104^{\circ}C\)

For part (B):
Initial temperature \(T_1 = 24^{\circ}C + 273=297\ K\)
Final temperature \(T_2=70^{\circ}C + 273 = 343\ K\)
Initial volume \(V_1 = 11.6\ L\)
Using Charles's law \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\), so \(V_2=\frac{V_1T_2}{T_1}\)
\(V_2=\frac{11.6\times343}{297}\ L\approx13.4\ L\)

Answer:

(A) \(104^{\circ}C\)
(B) \(13.4\ L\)