Question

An air bubble has a volume of 0.510 L at 18•C

1) What is the volume, in liters, at 5•C, if n and P do not change?

2) What is the volume, in liters, at 422 K, if n and P do not change?

3) What is the volume, in liters, at -10•C, if n and P do not change?

4) What is the volume, in liters, at 580 K, if n and P do not change?

Answer 1

Here we have to use Charles law, V/T=constant at constant pressure and amount of gas.

V1/T1=V2/T2 Given V1=0.510 L and T1=18°C=18+273=291 K,

(1) Here V2=? and T2=5°C=5+273=278 K

Therefore, V2=V1T2/T1

V2=(0.51 L x 278 K)/291 K

V2=0.487 L.

(2)

Here V2=? and T2=422 K

Therefore, V2=V1T2/T1

V2=(0.51 L x 422 K)/291 K

V2=0.0.7396 L.

(3)

Here V2=? and T2=-10°C=-10+273=263 K

Therefore, V2=V1T2/T1

V2=(0.51 L x 263 K)/291 K

V2=0.461 L.

(4)

Here V2=? and T2=580 K

Therefore, V2=V1T2/T1

V2=(0.51 L x 580 K)/291 K

V2=1.016 L.

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