Question

If the pressure of a gas sample is quadrupled and the absolute temperature is doubled, by what factor does the volume of the sample change?
A) 2

B) 1/4

C) 1/8

D) 1/2

Answer 1

Answer:

D) 1/2

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given,

P₂ = 4P₁

T₂ = 2T₁

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{4\times P_1}\times {V_2}}{2\times T_1}$

$V_2=\frac{1}{2}\times V_1$

The volume change by half of the original.

Answer 2

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:  

$\boxed{\boxed{ \ \frac{pV}{T} = constant \ }}$

• 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:

$\boxed{ \ \frac{p_2V_2}{T_2} = \frac{p_1V_1}{T_1} \ }$

Let us use the equation above to see the relationship between volumes. Enter all the information in the equation.

$\boxed{ \ \frac{4p_1V_2}{2T_1} = \frac{p_1V_1}{T_1} \ }$

$\boxed{ \ \frac{4V_2}{2} = V_1} \ }$

$\boxed{ \ 2V_2 = V_1} \ }$

$\boxed{ \ V_2 = \frac{1}{2}V_1} \ }$

Thus by factor ¹/₂, the volume of the sample will change.

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Notes

$\boxed{ \ \frac{pV}{nT} = R \ } \rightarrow \boxed{ \ pV = nRT \ }$

n = moles of ideal gas

R = the molar gas constant (in J mol⁻¹ K⁻¹)

Learn more

1. To what temperature would you need to heat the gas to double its pressure?  brainly.com/question/1615346#
2. The volume of Kr (in liters) brainly.com/question/6043528
3. The energy density of the stored energy 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