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8090 [49]
3 years ago
13

Round the following off to required number of significant figures.

Physics
2 answers:
Ray Of Light [21]3 years ago
6 0
140 but I’m not sure. Have a great day
Stells [14]3 years ago
4 0

Answer:

140

hope this helps

have a good day :)

Explanation:

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A balloon contains 2.3 mol of helium at 1.0 atm , initially at 240 ∘C. What's the initial volume? What's the volume after the ga
pashok25 [27]
A) initial volume
We can calculate the initial volume of the gas by using the ideal gas law:
p_i V_i = nRT_i
where
p_i=1.0 atm=1.01 \cdot 10^5 Pa is the initial pressure of the gas
V_i is the initial volume of the gas
n=2.3 mol is the number of moles
R=8.31 J/K mol is the gas constant
T_i=240^{\circ}C=513 K is the initial temperature of the gas

By re-arranging this equation, we can find V_i:
V_i =  \frac{nRT_i}{p_i} = \frac{(2.3 mol)(8.31 J/mol K)(513 K)}{1.01 \cdot 10^5 Pa}=0.097 m^3

2) Now the gas cools down to a temperature of
T_f = 14^{\circ}C=287 K
while the pressure is kept constant: p_f = p_i = 1.01 \cdot 10^5 Pa, so we can use again the ideal gas law to find the new volume of the gas
V_f =  \frac{nRT_f}{p_f}= \frac{(2.3 mol)(8.31 J/molK)(287 K)}{1.01 \cdot 10^5 Pa} = 0.054 m^3

3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:
W=p \Delta V= p(V_f -V_i)
by using the data we found at point 1) and 2), we find
W=p(V_f -V_i)=(1.01 \cdot 10^5 Pa)(0.054 m^3-0.097 m^3)=-4343 J
where the negative sign means the work is done by the surrounding on the gas.
5 0
3 years ago
Pls helppp. Is this right?
andreyandreev [35.5K]

Answer:

yes you are totally right

7 0
3 years ago
Read 2 more answers
A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?
ankoles [38]
Given required solution

M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
=5m/s*2.4s
=12m
so W=12*100
W=1200J
4 0
3 years ago
What does AC stand for in physics?
Svetradugi [14.3K]

Answer:

Alternating current

3 0
3 years ago
Read 2 more answers
Water in a tank is pressurized by air and pressure measured using a multi-fluid manometer. Determine the gage pressure of air in
sattari [20]

Answer:

The gauge pressure of air is 110 kpa

Explanation:

Atmospheric pressure, P_{atm} = 101 Kpa

P_{gauge} + \rho_w gh_1 + \rho_o gh_2 -\rho_{Hg} gh_3 =P_{atm}

P_{gauge}  = P_{atm} - \rho_w gh_1 - \rho_o gh_2 +\rho_{Hg} gh_3

where;

ρw is the density of water = 1000 kg/m³

ρo is the density of oil = 800 kg/m³

ρHg is the density of mercury = 13,600 kg/m³

g is acceleration due to gravity = 9.8 m/s²

P_{gauge}  = 101,000 - (1000* 9.8*0.2) - (800* 9.8*0.3) +(13,600* 9.8*0.46)\\\\P_{gauge}  = 101,000 - 1960 - 2352 + 13610.26\\\\P_{gauge}  = 110,298.26 pa

Therefore, the gauge pressure of air is 110 kpa

4 0
4 years ago
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