Question

An ideal monatomic gas expands isothermally from 0.540 m³ to 1.25 m³ at a constant temperature of 570 K. If the initial pressure is 1.20 ✕ 10⁵ Pa find the following.

(a) the work done on the gas
J

(b) the thermal energy transfer Q
J

(c) the change in the internal energy
J

Answer 1

Solution)

Part a)

We know,

Ideal gas equation,

Pv = nRT

1.20*10^5 Pa*(0.540)= n *8.314 * 570

n = 13.67

Now, work done

W = -nRTln (V2/V1) = -(13.67)*8.314 * 570 ln (1.25/0.540) =-54373.33 J

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Part b)

Here, change in internal energy, dU=0

So, thermal energy transfer, Q= 54373.33 J

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Part c)

Change in internal energy, dU=0

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