Question 977: Thermodynamics & KTG

Question

At 10
o
C the value of the density of a fixed mass of an ideal gas divided by it pressure is x. At 110
o
C this ratio is

Accepted Answer

PV = nRT
$$ \Rightarrow PV = {m \over M}RT$$
$$ \Rightarrow {{PV} \over m} = {{RT} \over M}$$
$$ \Rightarrow {P \over ho } = {{RT} \over M}$$
$$ \Rightarrow {P \over ho } \propto T$$
$$ 	herefore {P \over ho } \propto {1 \over T}$$
$${{{{{P_1}} \over {{ho _1}}}} \over {{{{P_2}} \over {{ho _2}}}}} = {{{T_1}} \over {{T_2}}} = {{383} \over {283}}$$
$$ \Rightarrow {x \over {{{{P_2}} \over {{P_2}}}}} = {{383} \over {283}}$$
$$ 	herefore {{{P_2}} \over {{P_2}}} = {{383} \over {283}}x$$

Answer

$${{10} \over {110}}x$$

Answer

$${{283} \over {383}}x$$

Answer

$$x$$

Answer

$${{383} \over {283}}x$$

Thermodynamics & KTG

At 10 o C the value of the density of a fixed mass of an ideal gas divided by it pressure is x. At 110 o C this ratio is

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