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In a thermodynamic process two moles of a monatomic ideal gas obeys P/V^2 = constant. If temperature of the gas increases from 300K to 400K, then find work done by the gas .?
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In a thermodynamic process two moles of a monatomic ideal gas obeys P/...
Answer:
Given:
Two moles of a monatomic ideal gas obeys P/V^2 = constant
Initial temperature, T1 = 300K
Final temperature, T2 = 400K
To find:
Work done by the gas
Solution:
As given, the gas obeys P/V^2 = constant.
Using the ideal gas law, PV = nRT
So, we can write P = nRT/V and substitute it in the given equation:
nRT/V * 1/V = constant
Therefore, T/V^2 = constant
As the temperature increases from T1 to T2, the volume also changes. Let V1 and V2 be the initial and final volumes respectively.
Using the above equation, we can write:
T1/V1^2 = T2/V2^2
Therefore, V2^2/V1^2 = T2/T1
So, V2/V1 = sqrt(T2/T1)
As we know that PV = nRT, we can write:
P1V1 = nRT1 and P2V2 = nRT2
Dividing the two equations, we get:
P2/P1 = T2/T1 * V1/V2^2
As P1/V1^2 = P2/V2^2, we can write:
V2/V1 = sqrt(P2/P1)
Substituting the value of V2/V1, we get:
sqrt(P2/P1) = sqrt(T2/T1)
Therefore, P2/P1 = T2/T1
As the gas is expanding, the work done by the gas is:
W = ∫P dV, where ∫ represents integration
W = ∫nRT/V dV
W = nRT ln(V2/V1)
Substituting the values of V2/V1 and nR, we get:
W = 2/3 * (T2 - T1)
Substituting the given values, we get:
W = 2/3 * (400 - 300) = 66.67 J
Conclusion:
The work done by the gas is 66.67 J.
Community Answer
In a thermodynamic process two moles of a monatomic ideal gas obeys P/...
554 J approx
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In a thermodynamic process two moles of a monatomic ideal gas obeys P/V^2 = constant. If temperature of the gas increases from 300K to 400K, then find work done by the gas .?
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