Respuesta :
Answer:
D) 1/2
Explanation:
Using Ideal gas equation for same mole of gas as
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]
Given,
P₂ = 4P₁
T₂ = 2T₁
Using above equation as:
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{4\times P_1}\times {V_2}}{2\times T_1}[/tex]
[tex]V_2=\frac{1}{2}\times V_1[/tex]
The volume change by half of the original.
¹/₂
Further explanation
Given:
- P₂ = 4P₁
- T₂ = 2T₁
Question:
By what factor does the volume of the sample change?
The Process:
We use an equation of state for an ideal gas:
[tex]\boxed{\boxed{ \ \frac{pV}{T} = constant \ }}[/tex]
- p = pressure (in Pa)
- V = volume (in m³)
- T = temperature (in Kelvin)
For the same amount of substances in two states, the equations for state-1 and state-2 are as follows:
[tex]\boxed{ \ \frac{p_2V_2}{T_2} = \frac{p_1V_1}{T_1} \ }[/tex]
Let us use the equation above to see the relationship between volumes. Enter all the information in the equation.
[tex]\boxed{ \ \frac{4p_1V_2}{2T_1} = \frac{p_1V_1}{T_1} \ }[/tex]
[tex]\boxed{ \ \frac{4V_2}{2} = V_1} \ }[/tex]
[tex]\boxed{ \ 2V_2 = V_1} \ }[/tex]
[tex]\boxed{ \ V_2 = \frac{1}{2}V_1} \ }[/tex]
Thus by factor ¹/₂, the volume of the sample will change.
- - - - - - - - - -
Notes
[tex]\boxed{ \ \frac{pV}{nT} = R \ } \rightarrow \boxed{ \ pV = nRT \ }[/tex]
n = moles of ideal gas
R = the molar gas constant (in J mol⁻¹ K⁻¹)
Learn more
- To what temperature would you need to heat the gas to double its pressure? https://brainly.com/question/1615346#
- The volume of Kr (in liters) https://brainly.com/question/6043528
- The energy density of the stored energy https://brainly.com/question/9617400
Keywords: the pressure of a gas sample, an ideal gas, volume, constant, moles, equation of state , quadrupled, the absolute temperature, doubled, by what factor, change
