a) Suppose you have a closed rigid container of air at 20 Celsius on the surface

Question

a) Suppose you have a closed rigid container of air at 20 Celsius on the surface of the Earth. We increase the temperature of the container to 77, leaving it closed, and read a pressure gauge attached to the container. Initially the reading was 101 KPa (kilopascals, or 1000 N/m2). What is the new reading? Give your answer in KPa.

b) The density of water is about 1000 kg/m3. Go below the surface of the ocean to a depth of 74 meters. What is the pressure at that depth? Use units of Earth's atmospheric pressure, 101 KPa (1 KPa = 101,000 N/m2) and include the contribution from Earth's atmosphere too (1 extra unit on this scale).

Explanation / Answer

Given
a)
T1 = 20 0C , T2 = 77 0C

P1 = 101 k Pa = 101*10^3 Pa

P2 = ?

   from ideal gas equation PV = nRT

   here V,n are consntant ==> P1/P2 = T1/T2

               P2 = P1*T2/T1 = 101000 *350.15/293.15 Pa = 120638.410 Pa = 120.64 kPa

the final gauge pressure is 120.64 k Pa

b)

we know that the pressure of the water increases as we go deep in the water

   P- p0 = rho*g*h

where h is the depth of the water and P0 is atmospheric pressure 1 atm = 1.01*10^5 Pa

   P = P0+ rho*g*h

   P = 1.01*10^5+1000*9.8*74 Pa

   P = 826200 Pa

   P = 8.262 *10^5 Pa

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