A 3.0 mol gas sample occupies 6.0 liters at 25 °C. What is the pressure of the gas in kPa’s?

Answer 1

The pressure will be quite high!

#P=(nRT)/V#

#=# #{(3.0*cancel(mol)*0.0821*cancel(L)*atm*cancel(K^(-1))cancel(mol)^(-1)xx298cancelK)/(6.0*cancelL)}xx101.3*kPa*(atm)^-1#

Would you anticipate that the pressure you obtain would be higher or lower than atmospheric pressure? Is the pressure you obtain reasonable?

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Answer 2

Olivia Anton

The pressure will be #"1240 kPa"#

Use the ideal gas law with the equation #PV=nRT#, where #P# is pressure, #V# is volume, #n# is moles, #R# is the gas constant, and #T# is the Kelvin temperature.

Given/Known #V="6.0 L"# #n="3.0 mol"# #R="8.3144598 L kPa K"^(-1) "mol"^(-1)"# https://tutor.hix.ai #T="25"^"o""C"+273.15="298 K"#

Unknown pressure, #P#

PV=nRT in the equation

Solution Rearrange the equation to isolate #P# and solve.

#P=(nRT)/V#

#P=(3.0cancel"mol"xx8.3144598cancel"L" "kPa" cancel("K"^(-1)) cancel("mol"^(-1))xx298cancel"K")/(6.0cancel"L")="1240 kPa"# (rounded to three significant figures) https://tutor.hix.ai

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Answer 3

Jose Ayala

The pressure of the gas is 150 kPa.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.