The volume of a gas–filled balloon is 30.0 liters at 313 K and 153 kilopascals pressure. What would the volume be at standard temperature and pressure (0°C and 100 kilopascals)?

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popcornrox37 popcornrox37 · Answer 1 · 2019-02-05T05:26:14+01:00

Use the combined and idea gas law to solve this:

[tex]\frac{P₁V₁}{T₁} = \frac{P₂V₂}{T₂}[/tex]

Let the subscript ₁ represent the initial values from pressure volume and temperature and ₂ for final.

So we know

P₁ = 153kpa

V₁ = 30L

T₁ = 313K

P₂ = 100kpa

T₂ = 0 Celsius which is 273 K

So we need to solve for volume:

[tex]T₂( \frac{P₁V₁}{T₁} )[/tex] = P₂V₂

Then we divide by Pressure on the right to give:

[tex]T₂( \frac{P₁V₁}{T₁} )[/tex] / P₂ = V₂

Plugging all that in:

[tex]273K( \frac{153kpa 30L}{313} )[/tex] / 100kpa = V₂

V₂ = 40.03 L

onyebuchinnaji onyebuchinnaji · Answer 2 · 2022-02-03T13:04:51+01:00

The volume of the gas at standard temperature and pressure is 40.03 L.

The volume of the gas at standard temperature and pressure is calculated as follows;

[tex]\frac{V_1P_1}{T_1} = \frac{V_2P_2}{T_2} \\\\V_1 = \frac{V_2P_2T_1}{T_2P_1}[/tex]

where;

[tex]V_1 = \frac{V_2P_2T_1}{T_2P_1}\\\\V _1 = \frac{30 \times 153 \times 273}{313 \times 100} \\\\V_1 = 40.03 \ L[/tex]

Thus, the volume of the gas at standard temperature and pressure is 40.03 L.

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