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If #3 L# of a gas at room temperature exerts a pressure of #7 kPa# on its container, what pressure will the gas exert if it the container's volume changes to #2 L#?

Answer 1

Emily Abbate

#P_2=10,5 k Pa#

#P_1*V_1=P_2*V_2# #7*3=P_2*2# #21=2*P_2# #P_2=21/2# #P_2=10,5 k Pa#

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Answer 2

Parker Abbott

Using the ideal gas law equation, (PV = nRT), where (P) is pressure, (V) is volume, (n) is the number of moles of gas, (R) is the ideal gas constant, and (T) is temperature in Kelvin, we can solve for the final pressure.

Given that the initial pressure (P_1 = 7 , \text{kPa}), the initial volume (V_1 = 3 , \text{L}), and the final volume (V_2 = 2 , \text{L}), and assuming constant temperature, we can set up the equation as follows:

[P_1V_1 = P_2V_2]

Substituting the given values:

[7 , \text{kPa} \times 3 , \text{L} = P_2 \times 2 , \text{L}]

[21 , \text{kPa} \cdot \text{L} = P_2 \cdot \text{L}]

[P_2 = \frac{21 , \text{kPa} \cdot \text{L}}{2 , \text{L}} = 10.5 , \text{kPa}]

Therefore, the gas will exert a pressure of (10.5 , \text{kPa}) when the container's volume changes to (2 , \text{L}).

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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.