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1.

About 0.014 kg nitrogen is enclosed in a vessel at temperature of `27^(@)C` How much heat has to be transferred to the gas to double the rms speed of its molecules ? `(R = 2 cal//mol K)`

Answer» As gas is enclosed in the cylinder, V = constant
`(Delta Q)_(v) = mu C_(v) Delta T`
Here, `mu = (0.014 xx 10^(3))//28 = (1//2) mol`
And as nitrogen is diatomic, `C_(v) = (5//2) R`,
Futher, as according to the given problem,
`((nu_(rms))_(2))/((nu_(rms))_(1)) = sqrt((T_(2))/(T_(1))) = 2`, i.e., `T_(2) = 4T_(1)`
`Delta T = 4 T_(1) - T_(1) = 3T_(1) = xx 300 = 900 K`
`(Delta Q)_(v) = (1)/(2) xx (5)/(2) xx 2 xx 900 = 2250 cal`
Discussion
2.

In the previous problem, the temperature of the isothemal process is how much lower the maximum temperature :A. `(P_(0) V_(0))/(nR)`B. `(P_(0) V_(0))/(2nR)`C. `(3P_(0) V_(0))/(2nR)`D. `(P_(0) V_(0))/(4nR)`

Answer» Correct Answer - D
d. From the last problem, from Eq. (ii), at point `A(V = V_(0))`
`T = (2 P_(0) V_(0))/(nR)`
This is the temperature which is constant along isothermal process
`T_(max)` is for` V = (3)/(2) V_(0)`
`T_(max) = (9 P_(0) V_(0))/(4 nR)` [from Eq. (iii)]
`T_(max) - T = (P_(0) V_(0))/(4nR)`
Discussion