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Question: In the temperature range 310 K to 330 K, The pressure P of a certain non ideal gas is related to volume V and temperature T byp=(24.9JK)TV-(0.00662JK2)T2V

How much work is done by the gas if its temperature is raised from 315 Kto 325 Kwhile the pressure is held constant?

Short Answer

Expert verified

Answer

Work done by the gas is 207 J.

Step by step solution

01

Step 1: Given

T1=315KT2=325KP=24.9TV-0.00662T2V

02

Determine the concept

Formula is as follow:

W=PΔV

Here, Pis pressure, V is volume and W is work done.

03

Determine the work done by the gas

Write the expression for pressure and volume as:

P=24.9TV-0.00662T2V∴V=24.9TP-0.00662T2P

Solve for the work done as:

W=Pâ–µVW=P24.9TP-0.00662T2PW=P24.9â–µTP-0.00662â–µ(T2)PW=P24.9T2-T1P-0.00662T22-T12P

Substitute the values and solve as:

W=P24.9325-315P-0.00662(3252-3152)PW=24.9325-315-0.00662(3252-3152)W=206.632≈207J

Hence, thework done by the gas is 207 J.

Therefore, the work done by the gas can be found by using the formula for work done in isobaric process.

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