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For many purposes we can treat nitrogen as an ideal gas at temperatures above its boiling point of - 196.°C.
Suppose the temperature of a sample of nitrogen gas is raised from -98.0 °C to -89.0 °C, and at the same time the pressure is increased by 10.0%.
A) Does the volume of the sample increase, decrease, or stay the same?

Answer :

Eduard22sly

Answer: the volume of the sample decreased

Explanation:

T1 = -98°C = - 98 + 273 = 175K

T2 = -89°C = -89 +273 = 184K

P1 = P

P2 = 110%P = 1.1P

V1 = V

V2 =?

P1V1/T1 = P2V2/T2

PxV/175 = 1.1PxV2/184

175x1.1PxV2 = PVx 184

V2 = (PVx 184) /(175x1.1P)

V2 = 0.96V = 96%V

Therefore, the final volume is 96% of the initial volume. This means that the final volume decreased by 96%

The volume of the sample will increase.  

• Based on the given information,  

• Let us assume that we have constant number of moles of nitrogen gas at -98 degree C, and the initial pressure is P1.  

• It is given that the pressure is increased by 10%, and the temperature is increased to -89 degree C.  

Now, the final pressure (P2) will be,  

P1 + P1*10/100 = 1.10 P1

T1 = -98 degree C = -98 + 273 K = 175 K

T2 = -89 degree C = -89 + 273 K = 184 K

At constant no of moles, the ideal gas equation is,  

PV = nRT

Here, n and R are constant, So, P1V1/T1 = P2V2/T2

P1 * V1/T1 / 175 K = 1.10 P1 * V2/184K

V2/V1 = 184/175 * 1.10 = 1.15  

V2 = 1.15 V1

Thus, the volume of sample increase by 1.15 times from the initial volume.

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