Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
Related questions
Question
I underlined the part I need help with it says, " V ↓ by __ %"
![... A gas sample initially in 1.31 L at 298 K was cooled to 273 K
under constant P. Find its new volume:
1.31 L/298 K = V2/273 K
V2 =
273 K 1.31 L 1,2001 L
%3D
%3D
%3D
298 K
A gas sample initially in 0.56 cm3 at 25 °C was heated to take up
5.6 L under constant P. Find its new toc:
T = (25 + 273) K = 298 K
0.56 cm3/29% K 5.6 L/T2
%3D
%3D
%3D
T, = 298 K:5.6 L x 1cm3 =2,980,000 K
10-3 L
%3D
%3D
0.56 cm3
toc,2 = (2980,00d - 273 ) °C =2,980,000
°C
%3D
%3D
What happens to a party balloon if you take it outside to 20 °F
const: P & n
from indoors at 72 °F?
T = 72 °F - 32 °F + 273.15 K = (72- 32) °F x 1K + 273.15 K =
295.37 K
%3D
%3D
%3D
1.8 °F
%3D
T2 = (20 - 32) °F x-1K + 273.15 K = 26618K
1.8 °F
%3D
%3D
45,37 K/V, = 2oc0.48K/V, V2 =In lobTSK =0902 = VI by %
V L29531K
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1cc13519-1c0b-438e-ad16-12cdea7ee242%2F840f73c8-55a8-4892-bb0b-08894f638705%2Fi033f2q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:... A gas sample initially in 1.31 L at 298 K was cooled to 273 K
under constant P. Find its new volume:
1.31 L/298 K = V2/273 K
V2 =
273 K 1.31 L 1,2001 L
%3D
%3D
%3D
298 K
A gas sample initially in 0.56 cm3 at 25 °C was heated to take up
5.6 L under constant P. Find its new toc:
T = (25 + 273) K = 298 K
0.56 cm3/29% K 5.6 L/T2
%3D
%3D
%3D
T, = 298 K:5.6 L x 1cm3 =2,980,000 K
10-3 L
%3D
%3D
0.56 cm3
toc,2 = (2980,00d - 273 ) °C =2,980,000
°C
%3D
%3D
What happens to a party balloon if you take it outside to 20 °F
const: P & n
from indoors at 72 °F?
T = 72 °F - 32 °F + 273.15 K = (72- 32) °F x 1K + 273.15 K =
295.37 K
%3D
%3D
%3D
1.8 °F
%3D
T2 = (20 - 32) °F x-1K + 273.15 K = 26618K
1.8 °F
%3D
%3D
45,37 K/V, = 2oc0.48K/V, V2 =In lobTSK =0902 = VI by %
V L29531K
%3D
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