Question 1:
Explain the existence of atmospheric pressure.
Answer:
Atmospheric pressure is the pressure due to the weight of the layer of air acting on the surface of the earth.
Explain the existence of atmospheric pressure.
Answer:
Atmospheric pressure is the pressure due to the weight of the layer of air acting on the surface of the earth.
Question 2:
The height of the mercury column in a barometer is 756 mm Hg on a cloudy day.
Calculate the atmospheric pressure at that time in pascal.
[Density of mercury, ρ = 13 600 kg m–3 and gravitational acceleration, g = 9.81 m s–2]
Answer:
$$ \begin{aligned} & h=756 \mathrm{~mm}=0.756 \mathrm{~m} \\ & P_{\mathrm{atm}}=? \\ & \rho=13600 \mathrm{~kg} \mathrm{~m}^{-3} \\ & g=9.81 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} $$
$$ \begin{aligned} &\text { Atmospheric pressure }\\ &\begin{aligned} & P_{a t m}=h \rho g \\ & P_{a t m}=(0.756) \times(13600) \times(9.81) \\ & P_{a t m}=100862 \mathrm{~Pa} \end{aligned} \end{aligned} $$
The height of the mercury column in a barometer is 756 mm Hg on a cloudy day.
Calculate the atmospheric pressure at that time in pascal.
[Density of mercury, ρ = 13 600 kg m–3 and gravitational acceleration, g = 9.81 m s–2]
Answer:
$$ \begin{aligned} & h=756 \mathrm{~mm}=0.756 \mathrm{~m} \\ & P_{\mathrm{atm}}=? \\ & \rho=13600 \mathrm{~kg} \mathrm{~m}^{-3} \\ & g=9.81 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} $$
$$ \begin{aligned} &\text { Atmospheric pressure }\\ &\begin{aligned} & P_{a t m}=h \rho g \\ & P_{a t m}=(0.756) \times(13600) \times(9.81) \\ & P_{a t m}=100862 \mathrm{~Pa} \end{aligned} \end{aligned} $$
Question 3:
What is the actual pressure at a depth of 125 m in a dam? State your answer in m H2O and pascal.
[Atmospheric pressure = 10.3 m H2O, density of water, ρ = 1 000 kg m–3 and gravitational acceleration, g = 9.81 m s–2]
Answer:
$$ \begin{aligned} & P=? \\ & h=125 \mathrm{~m} \\ & \rho=1000 \mathrm{~kg} \mathrm{~m}^{-3} \\ & P_{\text {atm }}=10.3 \mathrm{~m} \\ & g=9.81 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} $$
$$ \begin{aligned} & \text { Actual pressure (in unit } \mathrm{mH}_2 \mathrm{O} \text { ) } \\ & =\text { water pressure }+ \text { atmospheric pressure } \\ & =125+10.3 \\ & =135.3 \mathrm{~m} \mathrm{H}_2 \mathrm{O} \end{aligned} $$
$$ \begin{aligned} &\text { Actual pressure (in unit Pa) }\\ &\begin{aligned} & P=h \rho g \\ & P=(135.3) \times(1000) \times(9.81) \\ & =1327293 \mathrm{~Pa} \\ & \approx 1.33 \times 10^6 \mathrm{~Pa} \end{aligned} \end{aligned} $$
What is the actual pressure at a depth of 125 m in a dam? State your answer in m H2O and pascal.
[Atmospheric pressure = 10.3 m H2O, density of water, ρ = 1 000 kg m–3 and gravitational acceleration, g = 9.81 m s–2]
Answer:
$$ \begin{aligned} & P=? \\ & h=125 \mathrm{~m} \\ & \rho=1000 \mathrm{~kg} \mathrm{~m}^{-3} \\ & P_{\text {atm }}=10.3 \mathrm{~m} \\ & g=9.81 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} $$
$$ \begin{aligned} & \text { Actual pressure (in unit } \mathrm{mH}_2 \mathrm{O} \text { ) } \\ & =\text { water pressure }+ \text { atmospheric pressure } \\ & =125+10.3 \\ & =135.3 \mathrm{~m} \mathrm{H}_2 \mathrm{O} \end{aligned} $$
$$ \begin{aligned} &\text { Actual pressure (in unit Pa) }\\ &\begin{aligned} & P=h \rho g \\ & P=(135.3) \times(1000) \times(9.81) \\ & =1327293 \mathrm{~Pa} \\ & \approx 1.33 \times 10^6 \mathrm{~Pa} \end{aligned} \end{aligned} $$